cone penetration test design guide for state geotechnical

Cone Penetration TestDesign Guide for

State Geotechnical Engineers

Author David Saftner Report Number 2018-32

Date Published November 2018

Minnesota Department of Transportation Research Services amp Library

395 John Ireland Boulevard MS 330 St Paul Minnesota 55155-1899

mndotgovresearch

To request this document in an alternative format such as braille or large print call 651-366-4718 or 1-800-657-3774 (Greater Minnesota) or email your request to ADArequestdotstatemnus Please request at least one week in advance

Technical Report Documentation Page

1 Report No

MNRC 2018-32

2 3 Recipients Accession No

4 Title and Subtitle

Cone Penetration Test Design Guide for State Geotechnical

Engineers

5 Report Date

November 2018 6

7 Author(s)

Ryan Dagger David Saftner and Paul Mayne

8 Performing Organization Report No

9 Performing Organization Name and Address

Department of Civil Engineering University of Minnesota Duluth 1405 University Drive Duluth MN 55812

10 ProjectTaskWork Unit No

CTS2017022 11 Contract (C) or Grant (G) No

(C) 99008 (WO) 249

12 Sponsoring Organization Name and Address

Minnesota Department of Transportation Research Services amp Library 395 John Ireland Boulevard MS 330 St Paul Minnesota 55155-1899

13 Type of Report and Period Covered

Final Report 14 Sponsoring Agency Code

15 Supplementary Notes

httpmndotgovresearchreports2018201832pdf 16 Abstract (Limit 250 words)

The objectives of this project are focused on a new cone penetration testing (CPT) geotechnical design manual for highway and transportation applications based on recent research and innovation covering the period from 2000 to 2018 A step-by-step procedure is outlined on how to use CPT data in the analysis and design of common geotechnical tasks Previous manuals are either very outdated with information from 1970-1996 or not appropriately targeted to transportation works This design document introduces modern and recent advancements in CPT research not otherwise captured in legacy manuals from the 1990rsquos and earlier Examples and case studies are provided for each topic interpreted using CPT measures In the manual a step-by-step procedure is outlined on how to use CPT data in analysis and design for typical geotechnical practices These topics which are applicable both to state highways and local roads include bridge foundations (including shallow footings and deep foundations) and soil characterization (including determination of standard soil engineering properties)

17 Document AnalysisDescriptors

Cone penetrometers Geographic information systems Soils by

properties Bridge foundations Soil mechanics

18 Availability Statement

No restrictions Document available from

National Technical Information Services

Alexandria Virginia 22312

19 Security Class (this report)

Unclassified

20 Security Class (this page)

Unclassified

21 No of Pages

225

22 Price

CONE PENETRATION TEST DESIGN GUIDE FOR STATE

GEOTECHNICAL ENGINEERS

Prepared by

Ryan Dagger

David Saftner

Department of Civil Engineering

University of Minnesota Duluth

Paul W Mayne

School of Civil amp Environmental Engineering

Georgia Institute of Technology Atlanta

November 2018

Published by

Minnesota Department of Transportation

Research Services amp Library

395 John Ireland Boulevard MS 330

St Paul Minnesota 55155-1899

This report represents the results of research conducted by the authors and does not necessarily represent the views or policies

of the Minnesota Department of Transportation the University of Minnesota Duluth or the Georgia Institute of Technology

This report does not contain a standard or specified technique

The authors the Minnesota Department of Transportation the University of Minnesota Duluth and the Georgia Institute of

Technology do not endorse products or manufacturers Trade or manufacturersrsquo names appear herein solely because they are

considered essential to this report because they are considered essential to this report

TABLE OF CONTENTS

CHAPTER 1 INTRODUCTION1

CHAPTER 2 DIRECT CPT METHOD FOR SOIL CHARACTERIZATION 4

21 Introduction 4

22 Soil Unit Weight 6

23 CPT Material Index 7

231 Step 1 Normalized Sleeve Friction 7

232 Step 2 Iteration 8

24 Soil Behavior Type (SBT) 8

241 Step 1 8

242 Step 2 9

25 Effective Stress Friction Angle 9

26 Stress History 10

27 Lateral Stress Coefficient 11

28 Undrained Shear Strength 12

29 Ground Stiffness and Soil Moduli 12

210 Coefficient of Consolidation 13

211 Hydraulic Conductivity14

212 Example Problems 15

2121 Example 1 Direct CPT Methods for Geoparameters on Sands 15

2122 Example 2 Direct CPT Methods for Geoparameters on Clay 33

CHAPTER 3 DIRECT CPT METHOD FOR SHALLOW FOUNDATIONS 56

31 Procedure 56

311 Step 1 Estimating Footing Dimensions57

312 Step 2 Soil Characterization 58

313 Step 2a Foundation soil formation parameter59

314 Step 3 Soil elastic modulus and Poissonrsquos ratio 60

315 Step 4 Net cone tip resistance 60

316 Step 5 Bearing capacity of the soil 61

317 Step 6 Settlement61

318 Step 7 Final Check 61


321 Example 3 Direct CPT Method on Sands 62

CHAPTER 4 DIRECT CPT METHOD FOR DEEP FOUNDATIONS 70

41 Introduction70

42 Modified UniCone Method72

421 Step 172

422 Step 272

423 Step 373

424 Step 473

43 Axial Pile Displacements 73


441 Example 5 Direct CPT Methods Axial Pile Capacity74

REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES

Figure Geoparameters determined from CPT 1

Figure Conventional method for shallow foundation design compared to direct CPT method 2

Figure Direct CPT evaluation of axial pile capacity 3

Figure Soil unit weight from CPT sleeve friction 6

Figure Approximate ldquorules of thumbrdquo method using CPT sounding from Wakota Bridge MN 7

Figure SBT zones using CPT Ic 9

Figure Yield stress exponent compared to CPT material index 11

Figure k vs dissipation time for 50 consolidation (Mayne 2017) 15

Figure CPT data from Benton County Minnesota for example problem 1 16

Figure Soil layers using ldquorules of thumbrdquo17

Figure Soil layer 1 using SBT method 23




Figure CPT data from Minnesota for example problem 2 34






Figure Dissipation t50 data 53

Figure Direct CPT method for shallow foundations56

Figure Direct CPT method introduction 58

Figure Differentiation of porewater pressure measurement locations (Lunne et al 1997) 58

Figure 25 Foundation soil formation parameter hs versus CPT material index Ic (Mayne 2017) 60

Figure 26 Diagram of footing profiles for Example 362

Figure 27 CPT data from Northern Minnesota 63

Figure 28 Schematic of foundation design associated with CPT data 64

Figure 29 Soil type for example problem 367

Figure 30 Direct CPT evaluation of axial pile capacity 70

Figure 31 UniCone Method soil behavior type using CPT (Mayne 2017) 71

Figure 32 Modified UniCone Method soil behavior type using CPT (Mayne 2017) 71

Figure 33 Deep foundation end bearing pile diagram74

Figure 34 CPT data from Minnesota for example problem 5 75

Figure 35 Soil layers using ldquorules of thumbrdquo for pile capacity example using direct CPT method 76

Figure 36 Soil layer 1 using SBT method 81



Figure 39 Soil layering compared to pilings 85

LIST OF TABLES

Table 1 Geoparameters calculated directly from CPT 4

Table 2 Minor Geoparameters 5

Table 3 Soil type compared to exponent m 10

Table 4 Geoparameters evaluated for Case Example 117

Table 5 Geoparameters 35

Table 6 Bearing capacity defined by soil type61

Table 7 SCPT Results 62

CHAPTER 1 INTRODUCTION

The purpose of this manual is to provide an analysis of soil characterization shallow footings and deep

foundations using direct cone penetration testing (CPT) methods Geotechnical site characterization is

important for evaluating soil parameters that will be used in the analysis and design of foundations

retaining walls embankments and situations involving slope stability Common practice is to determine

these soil parameters through conventional lab and in-situ testing An alternative method uses CPT

readings of cone tip resistance (qt) sleeve friction ( fs) and pore pressure (u2) directly to determine these

parameters such as unit weight effective friction angle undrained shear strength and many others

(Figure 1) Direct CPT methods are provided for these parameters which will be used in the designs for

shallow and deep foundations

Figure 1 Geoparameters determined from CPT

Designing shallow foundations is typically done in a two-part process determining the bearing capacity

and expected settlement (commonly referred to as displacement) of the soil to approximate the

required size and shape of a foundation The older traditional methods are no longer required with

many approaches existing for using CPT directly in the design of shallow foundations Results from these

methods can provide a direct assessment of bearing capacity andor settlement A specific approach to

the direct method has been recommended and tested using a database of 166 full-scale field load tests

(Figure 2) A one-part process is used to scale the measured cone penetrometer readings (ie

measured cone tip resistance sleeve friction and pore water pressure) to obtain the bearing capacity

and settlement of the soil A step-by-step procedure has been created to transition from the CPT data to

bearing capacity with settlement accounted for The steps consist of estimating a design footing width

1

and length while using the process in the Soil Characterization section to determine soil parameters

directly from the CPT data

Figure 2 Conventional method for shallow foundation design compared to direct CPT method

There are upwards of 40 different direct CPT methods that have been developed over the past five

decades to determine the axial compression capacity of a piling foundation Earlier direct CPT methods

relied on hand-recorded information where mechanical-type CPT cone tip resistance data would be

collected at 20 cm intervals

Herein the method recommended for deep foundation design is the Modified UniCone method which

uses all three readings of the modern electronic piezocone penetrometer (CPTu) while addressing a

variety of pile foundation types (Figure 3) The modified UniCone Method is based on a total of 330 pile

load tests (three times the original UniCone database) that were associated with SCPTu data Two

computer software programs have been recommended for analyzing pile movements andor

settlements

2

Figure 3 Direct CPT evaluation of axial pile capacity

3

Symbol Parameter

γt Soil total unit weight

Ic CPT material index

SBT Soil behavior type (SBT)

ʹ σp Preconsolidation stress

YSR Yield stress ratio

ϕʹ Effective friction angle

Ko Lateral stress coefficient

su Undrained shear strength

Dʹ Constrained modulus

Eʹ Drained Youngrsquos modulus

CHAPTER 2 DIRECT CPT METHOD FOR SOIL

CHARACTERIZATION

21 INTRODUCTION

A direct CPT method for determining the value of each of various geoparameters shown in Table 1 is

provided The parameter Ic is used in the derivation of several sequential parameters This section of

the guide may be referred to as these parameters are used in calculations throughout the shallow

foundations and deep foundations sections

Table 1 Geoparameters calculated directly from CPT

4

Kʹ Bulk modulus

ks Subgrade reaction modulus

MR Resilient modulus

Gmax Small-strain shear modulus

k Coefficient of permeability

cv Coefficient of consolidation

Symbol Parameter Equation

ρt Mass Density ρt = γtga

where ga = 98 ms2

σvo Total Stress σvo asymp Σ (γti middot Δzi)

c Effective cohesion ʹ Empirical c asymp 003σp

In clays c asymp 01cu

Other minor parameters can be used in calculations of the geoparameters in Table 1 These minor parameters are provided in Table 2

Table 2 Minor Geoparameters

5

22 SOIL UNIT WEIGHT

The total soil unit weight can be estimated from CPT sleeve friction resistance as shown in Figure 4

(Mayne 2014) This method is not applicable to organic clays diatomaceous soils peats or sensitive

soils

Figure 4 Soil unit weight from CPT sleeve friction

Use Equation 1 to calculate the soils total unit weight

1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898

Since soil unit weight is required for determining most geoparameters (including Soil Behavior Type)

estimating the soil type of the layers using the ldquorules of thumbrdquo method is a good first step before

determining more precise layering (Figure 5) Once a unit weight is determined for each noticeable layer

change these results can be used in later calculations such as ldquoCPT Material Indexrdquo To use the ldquorules

of thumbrdquo method some helpful guidelines are to assume sands are identified when qt gt 725 psi and u2

asymp uo while the presence of intact clays are prevalent when qt lt 725 psi and u2 gt uo The magnitude of

porewater pressures help to indicate intact clays such as soft (u2 asymp 2middotuo) firm (u2 asymp 4middotuo) stiff (u2 asymp 8middotuo)

and hard (u2 asymp 20middotuo) Fissured overconsolidated clays tend to have negative u2 values such that u2 lt 0

6

Figure 5 Approximate ldquorules of thumbrdquo method using CPT sounding from Wakota Bridge MN

23 CPT MATERIAL INDEX

The development of the CPT material index Ic has improved the initial classification of soil types and

calculation of soil parameters as shown in Table 1 To calculate Ic follow steps 1 and 2

231 Step 1 Normalized Sleeve Friction

Calculate Fr using Equation 2 if not provided with the CPT data gathered

2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)

232 Step 2 Iteration

Iterate using Equations 3-5 to determine Ic by initially using n = 1 to calculate a starting value of Ic The

exponent n is soil-type dependent n = 1 (clays) n asymp 075 (silts) and n asymp 05 (sands) Iteration converges

quickly which is generally after the 3rd cycle

3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784

24 SOIL BEHAVIOR TYPE (SBT)

Typically soil samples are not taken when using CPT The soil types are then inferred from the qt fs and

u2 readings To determine the types of soil from CPT data follow steps 1 and 2

241 Step 1

To determine the soil layers from CPT results calculate Ic by following the steps under section ldquoCPT

Material Indexrdquo After Ic has been determined through all specified depths use Figure 6 to classify the

type of soil by comparing each Ic value to normalized CPT readings (Fr and Qtn) from Equations 2 and 3

8

Figure 6 SBT zones using CPT Ic

242 Step 2

To determine if any of the soil layers contain ldquosensitive clays and siltsrdquo from zone 1 or ldquovery stiff

overconsolidated (OC) soilrdquo from zones 8 and 9 use Equations 6 and 7 If any soil layers are found within

zone 1 by Equation 6 then caution should be taken as these clays are prone to instability collapses and

difficulties in construction performance Very stiff OC sands to clayey sands of zone 8 (15 lt Fr lt 45)

and very stiff OC clays to silts of zone 9 (Fr gt 45) can be identified by Equation 7

6 Equation 6 errata This is an exponential expression (see Figure 5)

119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)

7 119876119905119899 gt 0005(119865119903minus1)minus00003(119865119903minus1)2minus0002

1

Errata See terms and coefficients in Figure 5 above (numbers cannot be rounded off)

After each CPT reading has been assigned a zone from Figure 14 a visual representation can be made to

show the predominant layers by soil types (Figure A5)

25 EFFECTIVE STRESS FRICTION ANGLE

The effective friction angle (ϕ) is used to govern the strength for sands and clays where Equation 8 is

used for sands and Equation 9 is used for clays The value of ϕ for sands is derived from Ic so before ϕ

can be calculated refer to the iteration of Qtn under the ldquoCPT Material Indexrdquo section Once the values

9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072

Note m may be higher than 11 in fissured clays

of Ic and Qtn are calculated the type of soil can be determined from the section ldquoSoil Behavior Type

SBTrdquo The type of soil will dictate which equation to use for ϕ

Sands

8 120601prime(deg) = 176deg + 110deg log(119876119905119899)

Clays

9 119902119905minus120590119907119900 120601prime(deg) = 295deg ∙ 119861119902

0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900

Where = 119861119902 (1199062 minus 119906119900)(119902119905 minus 120590119907119900)

26 STRESS HISTORY

Determining the stress history can be characterized by an apparent yield stress ratio of the form

10 prime 120590119901

119884119878119877 = prime 120590119907119900

Where σp is defined as the preconsolidation stress or effective yield stress (Equation 10) The YSR is the

same equation as the more common overconsolidation ratio (OCR) but is now generalized to

accommodate mechanisms of preconsolidation such as ageing desiccation repeated cycles of wetting-

drying repeated freeze thaw cycles and other factors

11 prime prime 120590119901 = 120590119910 = 033(119902119905 minus 120590119907119900)119898prime

Where σp σy σvo and qt have units of kPa and the value of m depends on soil type with typical values shown in Table 3

Table 3 Soil type compared to exponent m

10

The value of m for non-fissured soils and inorganic clays and silts is derived from Ic (Figure 7) so before

m can be calculated (Equation 12) refer to the iteration of Ic under the ldquoCPT Material Indexrdquo section

Determine Ic for all soil layers before calculating m

12 028

119898prime = 1 minus 25 119868119888 1+( frasl ) 265

Once m is known for all soil layers the YSR can be determined

Figure 7 Yield stress exponent compared to CPT material index

A limiting value of YSR can be reached for clays and sands It can be calculated using Equation 13

13 (1 sin(emptyprime))

(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]

(1minussin(emptyprime))2

27 LATERAL STRESS COEFFICIENT

The lateral stress coefficient Ko = σhoσvo commonly referred to as the at-rest condition is used to

represent the horizontal geostatic state of soil stress Ko can be calculated using Equation 14 but

11

14

119870119900 = (1 minus sin(emptyprime)) ∙ 119884119878119877sin( emptyprime)

sections such as ldquoStress Historyrdquo and ldquoEffective Stress Friction Anglerdquo will need to be referred to for

determining parameters ϕ and YSR

A maximum value for Ko can be determined by Equation 15

15 (1+sin(emptyprime))

119870119900119898119886119909 = = 1199051198861198992(45deg + emptyprime2) (1minussin(emptyprime))

28 UNDRAINED SHEAR STRENGTH

Loading on soils can result in fully drained partially drained or fully undrained conditions Sands

typically produce drained cases due to their high permeability but exceptions may occur in loose sands

during fast loading where the water does not have sufficient time to dissipate Clays exhibit low

permeability and thus often result in undrained loading cases when a load is applied quickly For soft-

firm clays the undrained shear strength (su) can be determined from CPT via Equation 16 where the

value of the bearing factor Nkt can be taken as 12

16 119902119905minus120590119907119900 119904119906 =

119873119896119905

In the case of remolded undrained shear strength from CPT su asymp fs

29 GROUND STIFFNESS AND SOIL MODULI

Determining the grounds stiffness can be measured from geoparameters such as the constrained

modulus (D) drained Youngrsquos modulus (E) bulk modulus (K) subgrade reaction modulus (ks) resilient

modulus (MR) and small-strain shear modulus (Gmax)

17 119863 prime asymp 5 ∙ (119902119905 minus 120590119907119900)

18

12

119864 prime = 119863 prime

11

19 119864prime

119870prime = [3∙(1minus2119907prime)]

The subgrade modulus (ks) is a combination of soil-structural properties which creates a parameter that

depends on the ground stiffness and the size of the loaded element

20 119864prime

119896119904 = [119889∙(1minus1199072)]

The resilient modulus MR applies to pavement analysis and design and can be calculated using Equation

21 where qt and fs are in MPa

21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244

The small strain shear modulus Gmax is a representation of the initial stiffness of all soils and rocks

Graphically it is the beginning portion of all stress-strain-strength curves for geomaterials Use Equation

22 to determine Gmax

22 119866119898119886119909 = 120588119905 ∙ 1198811199042

Where Vs = shear wave velocity (Equation 23) as measured by seismic cone penetration tests (SCPT) If

only cone penetration tests (CPT) or piezocone (CPTu) data are available the shear wave velocity may

be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905

Where qt and fs have units of (kPa)

210 COEFFICIENT OF CONSOLIDATION

The coefficient of consolidation (cv) controls the rate that foundation and embankment settlements

occur By using results of CPT dissipation tests that measure the change in u2 readings over time the

value of cv can be determined Using Equation 24 cv can be determined based on piezocone dissipation

13

curves The equation below requires an estimate of the in-situ rigidity index (IR) of the soil If results of

SCPTU are available then IR may be determined from Gmax and qt per equation A38

24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550

Where ac = penetrometer radius (178 cm for 10-cm2 cone 220 cm for 15-cm2 cone) t50 = time to reach 50 dissipation IR = Gsu = undrained rigidity index G can be determined from the ldquoGround Stiffness and Soil Modulirdquo section

211 HYDRAULIC CONDUCTIVITY

The hydraulic conductivity also known as the coefficient of permeability (k) expresses the flow

characteristics of soils and has units of cms or feetday One method of calculation would be to use

Equation 25 where cv would need to be determined from ldquoCoefficient of Consolidationrdquo and D would

need to be determined from ldquoGround Stiffness and Soil Modulirdquo

25 119888119907∙120574119908 119896 =

119863prime

An alternative approach developed for soft normally-consolidated soils (Figure 8) is shown in Equation

26 where t50 (sec) values are used directly in assessing k in (cms)

26

14

125

119896 asymp ( 1

) 251∙11990550

Figure 8 k vs dissipation time for 50 consolidation (Mayne 2017)

212 EXAMPLE PROBLEMS

2121 Example 1 Direct CPT Methods for Geoparameters on Sands

Several geoparameters need to be determined based on the given CPT data collected for the South

Abutment of a bridge in Benton County MN (Figure 9) The groundwater table (GWT) was measured at

17 feet Determine all the geoparameters found in Table 4 at depths of 0 feet to 30 feet All sand layers

can be assumed ldquodrainedrdquo with ν = 02

15

Figure 9 CPT data from Benton County Minnesota for example problem 1

16


Symbol Parameter Symbol Parameter

γt Soil total unit weight Ko Lateral stress coefficient

Ic CPT material index Dʹ Constrained modulus

SBT Soil behavior type (SBT) Eʹ Drained Youngrsquos modulus ʹ σp Preconsolidation stress Kʹ Bulk modulus

YSR Yield stress ratio MR Resilient modulus

ϕʹ Effective friction angle Gmax Small-strain shear modulus

Solution

Soil total unit weight

Estimate soil layering using ldquorules of thumbrdquo

Figure 10 Soil layers using ldquorules of thumbrdquo

γw = 6224 pcf

17

Layer 1

From Figure 10 fs = 17 psi taken as a representative value of the layer

17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1

From Figure 10 fs = 17 psi and qc = 3500 psi

18

qt = qc + u2 ∙ (1 minus a) = 3500 psi + 3 psi ∙ (1 minus 08) = 3501 psi

= γt ∙ 6 feet = 1204 pcf ∙ 6 feet = 7225 psf = 50 psi σvo

fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049

(qt minus σvo) (3501 psi minus 50 psi)

Step 2 Iterate to solve for Ic Steps are not shown for brevity

uo = 0 psi

prime σvo = σvo minus uo = 5 psi minus 0 psi = 50 psi

(qt minus σvo)σatm (3501 psi minus 50 psi )145 psi = = Qtn prime )n (σvoσatm (50 psi145 psi)n

prime σvo 50 psi n = 0381 ∙ Ic + 005 ( ) minus 015 = 0381 ∙ Ic + 005 ( ) minus 015

σatm 145 psi

Ic = radic[347 minus log(Qtn)]2 + [122 + log(049)]2

Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19


= 7225 psf + γt ∙ 6 feet = 7225 psf + 1204 pcf ∙ 6 feet = 1445 psf = 100 psi σvo

fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081



uo = γwater ∙ (z minus 119911119908) = 6224 pcf ∙ (23 feet minus 17 feet) = 3734 psf = 99 psi




σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19

Soil Behavior Type (SBT)

Layer 1

Based on values of Ic Qtn and Fr the first layer is defined as a ldquoDrained Gravelly Sandrdquo from

Figure 11

Figure 11 Soil layer 1 using SBT method

23

Layer 2

Based on values of Ic Qtn and Fr the second layer is defined as a ldquoDrained Sandrdquo from Figure

12


24

Layer 3

Based on values of Ic Qtn and Fr the third layer is defined as a ldquoDrained Sandrdquo from Figure 13


25

Layer 4

Based on values of Ic Qtn and Fr the fourth layer is defined as a ldquoDrained Sandrdquo from Figure 14


Effective Stress Friction Angle

Layer 1

ϕprime(deg) = 176deg + 110deg log(Qtn) = 176deg + 110deg log(3533) = 456deg

Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028

prime m = 1 minus = 1 minus = 072 25 25 1 + (

Icfrasl ) 1 + (13frasl ) 265 265

prime prime σp = σy = 033(qt minus σvo)mprime = 033(24139 kPa minus 345 kPa)072 = 4717 kPa

= 684 psi

prime σp 684 psi YSR = = = 136

primeσvo 50 psi

Layer 2

028 028

prime m = 1 minus = 1 minus = 072 25 25 Ic 1 + (14 1 + ( frasl ) frasl ) 265 265


= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265

)mprime prime prime σp = σy = 033(qt minus σvo = 033(8274 kPa minus 168 kPa)072 = 215 kPa = 312 psi


primeσvo 188 psi

Lateral Stress Coefficient

Layer 1

Ko = (1 minus sin(emptyprime)) ∙ YSRsin( emptyprime) = (1 minus sin(456deg)) ∙ 136sin( 456deg) = 18

Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11



Dprime asymp 5 ∙ (qt minus σvo) = 5 ∙ (3501 psi minus 50 psi) = 17480 psi

Eprime 15890 psi

Kprime = = = 8828 psi [3 ∙ (1 minus 2vprime)] [3 ∙ (1 minus 2(02))]

053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4

Ground Stiffness and Soil Moduli

Layer 1

Values of qt and fs need to be in MPa

MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03

Vs (mfrasls) = [101 ∙ log(qt) minus 114]167 ∙ (100 ∙ ) qt

Values of qt and fs need to be in kPa

03 1172 kPa m Vs (mfrasls) = [101 ∙ log(24139 kPa) minus 114]167 ∙ (100 ∙ ) = 2745

24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37

ga 322 ftfrasls2 ft3

2 ft Gmax = ρt ∙ Vs

2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2


Dprime 17480 psi Eprime = = = 15890 psi

11 11

Eprime 15890 psi Kprime = = = 8828 psi

[3 ∙ (1 minus 2vprime)] [3 ∙ (1 minus 2(02))]

053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32

2122 Example 2 Direct CPT Methods for Geoparameters on Clay


Abutment (Figure 15) The groundwater table (GWT) was measured at 60 feet Determine all the

geoparameters found in Table 5 at depths of 0 feet to 42 feet All sand layers can be assumed ldquodrainedrdquo

ν = 02 All clay layers can assume ν = 049

33

Figure 15 CPT data from Minnesota for example problem 2

34

Table 5 Geoparameters


γt Soil total unit weight Eʹ Drained Youngrsquos modulus

Ic CPT material index Kʹ Bulk modulus

SBT Soil behavior type (SBT) MR Resilient modulus

ʹ σp Preconsolidation stress Gmax Small-strain shear modulus

YSR Yield stress ratio su Undrained shear strength

ϕʹ Effective friction angle cv Coefficient of consolidation

Ko Lateral stress coefficient k Hydraulic conductivity


Solution



35


Unit weight of water γw = 6224 pcf

Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1



σvo = γt ∙ 2 feet = 1179 pcf ∙ 2 feet = 235 psf = 16 psi

fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi


(qt minus σvo)σatm (3000 psi minus 16 psi )145 psi = = Qtn prime )n (16 psi145 psi)n (σvoσatm


σatm 145 psi


37

Qtn = 4126 n = 032 lt 10 Ic = 121 therefore sand

Layer 2



σvo = 235 psf + γt ∙ 30 feet = 235 psf + 1172 pcf ∙ 30 feet = 3751 psf = 260 psi

fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53

(q minus σ ) (252 psi minus 260 psi) t vo


uo = 0 psi


(qt minus σvo)σatm (250 psi minus 260 psi )145 psi = = Qtn prime (σvoσatm)n (260 psi145 psi)n

38


σatm 145 psi


Qtn = 87 n = 10 le 10 Ic = 32 therefore clay

Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi


(qt minus σvo)σatm (30016 minus 277)145 psi = = Qtn prime (σvoσatm)n (277 psi145 psi)n

39

primeσvo 277 psi n = 0381 ∙ Ic + 005 ( ) minus 015 = 0381 ∙ Ic + 005 ( ) minus 015

σatm 145 psi


Qtn = 1962 n = 052 le 10 Ic = 15 therefore sand

Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi


Qtn = 65 n = 10 lt 10 Ic = 29 therefore clayey silt


Layer 1

Based on values of Ic Qtn and Fr the first layer is defined as a ldquoDrained Sandrdquo from Figure 17

41


Layer 2

Based on values of Ic Qtn and Fr the second layer is defined as a ldquoUndrained Clayrdquo from Figure

18

42


43

Layer 3



44

Layer 4

Based on values of Ic Qtn and Fr the fourth layer is defined as a ldquoUndrained Silty Mixrdquo from

Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo

ϕprime(deg) = 295deg ∙ Bq0121 ∙ [0256 + 0336 ∙ Bq + log ( )]

primeσvo

Where

(u2 minus uo) (10 psi minus 0 psi) Bq = = = 004


252 psi minus 260 psi ϕprime(deg) = 295deg ∙ 00440121 ∙ [0256 + 0336 ∙ 0044 + log ( )]

260 psi

ϕprime(deg) = 245deg

Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265

prime prime σp = σy = 033(qt minus σvo)mprime = 033(3000 psi minus 33)072 = 1051 psi


primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028

prime m = 1 minus = 1 minus = 072 25 25 Ic 1 + ( frasl ) 1 + (15frasl ) 265 265

)mprime prime prime σp = σy = 033(qt minus σvo = 033(4002 psi minus 277)072 = 1288 psi

47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37


2 ft 2 Gmax = ρt ∙ Vs = 37 ∙ (8412 ) = 260 ∙ 106psf = 17992 psi

s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts

γt 121 pcf slug ρt = = = 38



2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s

Undrained Shear Strength

Layer 2

qt minus σv0 250 psi minus 26 psi su = = = 187 psi

12 12

Layer 4


12 12

Coefficient of Consolidation

52

Example dissipation data are shown in Figure 21 Example calculations are provided

Figure 21 Dissipation t50 data

Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50

Hydraulic Conductivity

Layer 1

cm 1 psi 359 ∙ 6224 pcf ∙ cv ∙ γw sec 144 pcf

k = = = 10 ∙ 10minus4 cmsec Dprime 14984 psi

Layer 2



Layer 3



54

Layer 4



55

CHAPTER 3 DIRECT CPT METHOD FOR SHALLOW

FOUNDATIONS

31 PROCEDURE

Shallow foundation analysis is typically done in a two-part traditional procedure The traditional

techniques are no longer required as a direct CPT method for square rectangular and circular shallow

footings is available (Figure 22) This process has the soil types grouped into four main categories sands

silts fissured clays and intact clays When determining soil types for each design it is believed footings

on sands and silts act in a fully drained manner while intact clays act in an undrained manner under

conditions of constant volume In order to determine the vertical stress-displacement-capacity of

square rectangular and circular shallow footings follow the steps provided towards the solution given

by Equation 27

Figure 22 Direct CPT method for shallow foundations

Equation 27 may be used to calculate all footing stresses from zero to the bearing capacity (qmax) To

calculate qmax for a sized footing of width (B) length (L) and thickness (t) follow steps 1 through 7

provided The settlement (s) can be determined after the calculation of qmax by simply rearranging

56

Equation 27 where the allowable stress (qallow) is defined as qmax divided by the factor of safety (FS) For

shallow footings a FS value of 3 is commonly used in geotechnical engineering

120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905

Where hs = the foundation soil formation parameter qtnet = the net corrected cone tip resistance

311 Step 1 Estimating Footing Dimensions

In Minnesota frost heave can have devastating effect on a shallow foundations It is common practice to

place a foundation bearing elevation below the expected maximum frost depth (roughly 45 to 6 feet

below ground level) With this assumption estimate a footing size (B x L) for design to obtain

representative data from CPT roughly 15middotB below the foundation depth (Df) as shown in Figure 23 The

CPT data collected will consist of

cone tip resistance (qc)

measured porewater pressure acting behind the cone tip (u2) as shown in Figure 24

sleeve friction (fs)

57

Figure 23 Direct CPT method introduction

Figure 24 Differentiation of porewater pressure measurement locations (Lunne et al 1997)

312 Step 2 Soil Characterization

The soil behavior type (SBT) and CPT material index (Ic) govern the value of the formation factor hs The

first step is following the steps in Soil Unit Weight to determine soil layering and γt values for each

layer To determine a representative unit weight (γsoil) for all the layers in the range of Df to 15∙B below

58

Df the user will need to use their engineering judgement on the definition of ldquorepresentative unit

weightrdquo This single value of unit weight is used in further calculations such as the total vertical soil stress (σvo) from Equation 29 and effective vertical stress (σvo) from Equation 30 Values of σvo and σvo

will need to be calculated at 15middotB below Df

120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30

Where z = Df +15middotB uo=γwatermiddot(z-zw)

The qc will need to be corrected using Equation 31 in the case of fine-grained soils that develop excess porewater pressure during cone penetration These values will be used in the following calculations

119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31

Where 119860119899 a = cone area ratio = eg MnDOT commonly uses a = 08 119860119888

119860119899 = cross-sectional area of load cell or shaft 119860119888 = projected area of the cone

The cone area ratio is determined based on the type of piezocone tip used during in-situ field testing

Manufacturer specifications should provide the measured net area ratio (a) for the particular cone

penetrometer as determined by calibration in a pressurized triaxial chamber

313 Step 2a Foundation soil formation parameter

With the representative value γsoil continue to follow the steps in CPT Material Index through Soil

Behavior Type (SBT) to better define the type of soil at depth 15∙B below Df Use the value of Ic

calculated at depth 15∙B below Df to calculate hs with Equation 36 This parameter is based on the soil

type with typical values shown in Figure 25 The data on silts and sands are considered fully drained

whereas the fissured clay subset may be partially drained to undrained

59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24

Figure 25 Foundation soil formation parameter hs versus CPT material index Ic (Mayne 2017)

314 Step 3 Soil elastic modulus and Poissonrsquos ratio

Details about the soil such as its elastic modulus (Es) and Poissonrsquos ratio (ν) will be needed for further

calculations A representative value for the Es can be determined from in-situ field tests Values of ν can

be assigned as 02 for drained sands and as 05 for undrained loading cases involving clays (Jardine et al

1985 Burland 1989)

315 Step 4 Net cone tip resistance

Calculate qtnet the mean value of net cone tip resistance 15middotB below the foundation bearing elevation

using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60

316 Step 5 Bearing capacity of the soil

Use the assumed B and L calculated hs and qtnet to determine the soils bearing capacity (qmax) from

Equation 34 Use Table 6 to calculate qmax by using the maximum allowable settlement ratio (sB)max

correlating to the soil type If hs is in between the given values interpolate to acquire (sB)max

Table 6 Bearing capacity defined by soil type

Type of Soil hs (sB)max

Clean Sands 058 12

Silts 112 10

Fissured Clays 147 7

Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909

317 Step 6 Settlement

Settlement can be calculated directly using the results from Equation 35 A FS equal to 3 is common in

foundation engineering

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905

318 Step 7 Final Check

Check that the applied stress (q) is less than qmax using Equation 36 Repeat the process again with a

new B andor new L if q gt qmax

05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS

321 Example 3 Direct CPT Method on Sands

A footing size needs to be determined based on the given CPT data collected for the South Abutment

(Figure 27) The footing stress (q) was determined to be 8000 psf Estimate a footing size (B x L) and

determine the bearing capacity of the foundation (qmax) using the direct CPT method provided Also

determine the expected settlement based on the calculated bearing capacity


The soil elastic modulus determined from seismic CPT (SCPT) are shown in Table 7

Table 7 SCPT Results

Depth (feet) Es (tsf)

Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62

Figure 27 CPT data from Northern Minnesota

Solution

Step 1 Assume the footing is placed below the frost depth of 6 feet

Estimate L L = 50 feet = 600 inches

63

Estimate B B = 12 feet = 144 inches

Estimate footing thickness t t = 2 feet

Df = 6 feet = 72 inches

Df + 15middotB = 24 feet = 288 inches

Step 2 Soil total unit weight


Figure 28 Schematic of foundation design associated with CPT data

Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi

Using the unit weights calculated between Df and 15B determine a representative unit weight of the soil to calculate the total and effective soil stresses Based on layer 3 being the weakest supporting layer γsoil = 113 pcf

σvo = γsoil ∙ (119863119891 + 15 ∙ B) = 113 lbsfraslft3 ∙ 24 feet = 2721 psf = 189 psi

prime σvo = σvo minus uo = 2721 psf minus 0 psf = 2721 psf = 189 psi

Calculate the cone tip resistance between Df and Df + 15B below the foundation depth


Step 2a CPT Material Index

Using Figure 28 the sleeve friction at 15B below the foundation depth is about 16 psi

65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13


Iterate to solve for Ic Steps are not shown for brevity



σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66


Use the Ic value to determine the foundation soil formation parameter

Calculate the foundation soil formation parameter The tip resistance is about 1250 psi at 24 feet and

the cone area ratio was determined to be 08 from information provided by the manufacturer

23 23 hs = 28 minus = 28 minus = 078 = SandSilt 15 15 Ic 210

1 + ( ) 1 + ( ) 24 24

67

Step 3 Determine representative values of soil elastic modulus from soil testing and Poissons ratio The

soil elastic modulus was taken as the average value between depths of Df and 15middotB (6 feet and 24 feet)

433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5

ldquoDrained sandsiltrdquo gives a ν = 020

Step 4 Calculate the net cone tip resistance

qtnet = qt minus σvo = 1250 psi minus 189 psi = 12311 psi

Step 5 Calculate the bearing capacity of the sand

In drained sandssilts the bearing capacity is taken as the stress when (sB) = 011 (or 11 foundation width)

minus0345 minus0345 05 s L 600 qmax = hs ∙ qtnet ∙ ( ) ∙ ( ) = 078 ∙ (9795) ∙ (011)05 ∙ ( )

B B 144

= 193 psi = 27860 psf qmax

Assuming a factor of safety (FS) of 3

qmax = 645 psi = 9287 psf

3

Step 6 Calculate settlement

68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B

2 0345 1 645 psi 600 inches s = 144 inches ∙ [ ∙ ∙ ( ) ] = 18 inches

078 12311 psi 144 inches

Step 7 Determine if q gt qmax

q = 8000 psf lt 9287psf = qmax3

69

CHAPTER 4 DIRECT CPT METHOD FOR DEEP FOUNDATIONS

41 INTRODUCTION

The axial compression capacity (Qtotal) for a single pile includes a side component (Qside) end bearing

component (Qbase) and pile weight (Wpile) as shown in Figure 30 Since piles commonly push through

several layers a summation of the unit side frictions acting on the pile segments must be considered

over the length of the pile While the examples shown in this Guide resemble hand calculations

computer software is more efficient Programming the procedure is possible and represents a practical

method of designing deep foundations However commercial software is frequently available and is

MnDOTrsquos most common method of designing deep foundations


decades to determine a piles axial compression capacity Many of the earliest methods relied on hand-

recorded information where qc data from mechanical CPTs would be collected at 20 cm intervals

whereas the direct CPT method uses scaled penetrometer readings via specified algorithms to obtain

the pile unit side friction and unit end bearing The method that will be used for deep foundation design

is the Modified UniCone method which uses all three readings of the electronic piezocone (qt fs and u2)

while addressing a variety of pile foundation types


70

The modified UniCone Method is based upon a total of 330 pile load tests (three times the original

Unicone database) that were associated with SCPTu data Originally the UniCone method provided

approximate soil classification in five groups via a chart of qE vs fs (Figure 31) where qE = qt -u2 Later

using the modified approach with a larger data set provided soil sub classifications as shown in Figure

32 This new 9-zone normalized soil behavior type is determined using CPT data in combination with the

CPT Material Index

Figure 31 UniCone Method soil behavior type using CPT (Mayne 2017)

Figure 32 Modified UniCone Method soil behavior type using CPT (Mayne 2017)

71

42 MODIFIED UNICONE METHOD

In order to determine the axial pile capacity using the modified method the first step requires the

determination of geoparameters as shown in Direct CPT Method for Soil Characterization Once the

soil unit weight and CPT Material index are determined for each soil layer then the effective cone

resistance pile unit side friction (fp) and pile end bearing resistance (qb) can be determined

421 Step 1

Work through the steps provided in CPT Method for Soil Characterization until each soil layer is

defined by its CPT material index and soil behavior type using Figure 6 After these steps have been

completed continue to Step 2 to determine qE qb and fp

422 Step 2

Once qt is determined for each layer the effective cone resistance can be calculated using Equation 37

119902119864 = 119902119905 minus 1199062 37

Where (a) qE is the specific value at each elevation along the pile sides for determining fp and (b) at the bottom of the pile qE is averaged in the vicinity of the pile tip from the tip bearing elevation to about one diameter beneath the tip for determining qb

Using CPT material index and qE the pile end bearing resistance is calculated using Equation 38

38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)

The pile unit side friction is obtained from qE and Ic

39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)

Where θPT = coefficient for pile type (084 for bored 102 for jacked 113 for driven piles) θTC = coefficient for loading direction (111 for compression and 085 for tension) θRATE = rate coefficient applied to soils in SBT zone 1 through 7 (109 for constant rate of penetration test and 097 for maintained load tests)

72

423 Step 3

Due to piles extending through multiple layers the unit side components acting on various pile

segments would need to be summed if not using the direct CPT method Since CPT calculates data at

regular intervals of 2 cm to 5 cm along the sides of the pile the average fp in each layer can be used

directly in Equation 40 to obtain the shaft capacity

119876119904119894119889119890 = 119891119901 ∙ 119860119904 40

Where fp = average pile side friction along pile length from eqn 39 As = πmiddotdmiddotH d = pile diameter and H = length embedded below grade

The base capacity for a pile in compression loading is given by Equation 41 For piles in tension (or uplift)

Qbase can be taken as 0

= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where

qb = end bearing resistance from eqn 38 Ab = πmiddotd24 (area of a circular pile)

424 Step 4

The final step is to calculate the axial pile capacity using Equation 42

119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42

43 AXIAL PILE DISPLACEMENTS

Movement of pile foundations can be assessed using elastic continuum theory which has been

developed using finite element analyses boundary elements and analytical closed-form solutions In

the case of piles passing through several soil layers the elastic solution can be used by stacking pile

segments (each with its own stiffness) as represented by soil Youngrsquos modulus The use of software is

recommended for pile groups Several available programs such as DEFPIG GROUP and PIGLET can

handle pile groups under axial and lateralmoment loading

73

44 EXAMPLE PROBLEMS

441 Example 5 Direct CPT Methods Axial Pile Capacity

Several piles need to be placed beneath the edge of a building Use the CPT data collected for this site

shown in Figure 34 to determine the axial capacity for one of the piles Assume round steel driven piles

will be used with a diameter of 1275 inches wall thickness of 025 inches and lengths of 80 feet

Concrete will be used to fill the piles To solve for all geoparameters use the Direct CPT Method for Soil

Characterization section All sand layers can be assumed ldquodrainedrdquo with ν = 02

Figure 33 Deep foundation end bearing pile diagram

74


75

Solution



Figure 35 Soil layers using ldquorules of thumbrdquo for pile capacity example using direct CPT method

Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1



∙ 4 feet = 1219 pcf ∙ 4 feet = 4877 psf = 34 psi σvo = γt

fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi


Qtn = 117 n = 10 le 10 Ic = 29 (ie clayey silt)

Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1

Based on values of Ic (14) Qtn (359) and Fr (06) the first layer is defined as a ldquoGravelly Sandrdquo

from Figure 36

80


81

Layer 2

Based on values of Ic (29) Qtn (117) and Fr (26) the second layer is defined as a ldquoSilty Mixrdquo

from Figure 37


82

Layer 3

Based on values of Ic (15) Qtn (180) and Fr (04) the third layer is defined as a ldquoSandrdquo from

Figure 38


Effective Cone Resistance

Layer 1

83

qE = qt minus u2 = 3004 psi minus 20 psi = 2984 psi

Layer 2


Layer 3


Pile End Bearing Resistance

Layer 3

= qE ∙ 10(0325∙Icminus1218) qb = 5000 psi ∙ 10(0325∙15minus1218) = 9085 psi

Pile Unit Side Friction

Layer 1

minus3605) fp = qE ∙ θPT ∙ θTC ∙ θRATE ∙ 10(0732∙Ic

fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity

The side capacity is based on pile design as shown in Figure 39

Layer 1

Qside = fp ∙ As = 99 psi ∙ (π ∙ d ∙ L) = 99 psi ∙ (π ∙ 1275 in ∙ 48 in) = 19064 lb

Layer 2


Layer 3


Figure 39 Soil layering compared to pilings

85

End Bearing Resistance in Layer 3

d2 1275 in2

Qbase = qb ∙ Ab = 9085 psi ∙ (π ∙ ) = 9085 psi ∙ (π ∙ ) = 115932 lb 4 4

Wp = (γsteel ∙ Asteel ∙ Depth) + (γsteel ∙ Aconc ∙ Depth)

Wp = (490 pcf ∙ 003 ft2 ∙ 55 ft) + (150 pcf ∙ 085 ft2 ∙ 55 ft) = 7724 lbs

Layer 3

Qtotal = ΣQside + Qblayer3 minus Wp

Qtotal = (19064 + 45713 + 58170) lbs + 115932 lbs minus 7724 lbs = 642563 lbs

= 642 kips Qtotal

Table 8 Summary of selected and calculated parameters

Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

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87

Burland J B (1973) Shaft Friction of Piles in Clay -A Simple Fundamental Approach Ground Eng 6(3) 30-42

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Chen BS-Y and Mayne PW (1996) Statistical relationships between piezocone measurements and stress history of clays Canadian Geotechnical Journal 33 (3) 488-498

Cho W and Finno RJ (2010_ Stress-strain responses of block samples of compressible Chicago glacial clays Journal of Geotechnical amp Geoenvironmental Engineering 136 (1) 178-188

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Clausen CJF Aas PM and Karlsrud K (2005) Bearing capacity of driven piles in sand the NGI approach Frontiers in Offshore Geotechnics (Proc ISFOG Perth) London Taylor amp Francis Group

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Eslaamizaad S and Robertson PK (1996) Cone penetration test to evaluate bearing capacity of foundations in sand 429-438 Proc CGCFG-49 Vol 1 St Johns Newfoundland

Eslami A Alimirzaei M Aflaki E and Molaabasi H (2017) Deltaic soil behavior classification using CPTu records Marine Georesources amp Geotechnology 35 (1) 62-79

88

Eslami A and Gholami M (2005) Bearing capacity analysis of shallow foundations from CPT data 1463-1466 Proc ICSMGE- 16 Vol 3 (Osaka) Millpress Rotterdam

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Finno R J and Chung C K (1992) Stress-strain-strength responses of compressible Chicago glacial clays Journal of Geotechnical Engineering 118 (10) 1607ndash1625

Fleming K Weltman A Randolph M and Elson K (2009) Piling Engineering Third Edition London Taylor amp Francis Group

Frank R and Magnan J-P (1995) Cone penetration testing in France national report Proceedings International Symposium on Cone Penetration Testing 3 (CPT95 Linkoumlping) Swedish Geotechnical Society Report 3(95) 147-156

Gavin K Adekunte A and OKelly B (2009) A field investigation of vertical footing response on sand Geotechnical Engineering 162 257-267

Hegazy YA (1998) Delineating geostratigraphy by cluster analysis of piezocone data PhD dissertation School of Civil amp Environmental Engineering Georgia Institute of Technology Atlanta GA

Holtz RD Kovacs WD and Sheehan TC (2011) An Introduction to Geotechnical Engineering 2nd Edition London Pearson Publishing

Hunt CE Pestana JM Bray JD and Riemer M (2002) Effect of pile driving on static and dynamic properties of soft clay Journal of Geotechnical amp Geoenvironmental Engineering 128 (1) 13-24

Jamiolkowski M Ladd CC Germaine JT and Lancellotta R (1985) New developments in field and lab testing of soils Proc ICSMFE-11 1 57-154

Jardine R Chow F Overy R and Standing J (2005) ICP design methods for driven piles in sands and clays London Thomas Telford Publishing

Jardine RJ Lehane BM Smith P and Gildea P (1995) Vertical loading experiments on rigid pad foundations at Bothkennar Geotechnique 45 (4) 573-597

89

Karlsrud K (2012) Prediction of load-displacement behavior and capacity of axially-loaded piles in clay based on analyses and interpretation of pile load test results PhD Dissertation Norwegian Univ Science amp Technology Trondheim Norway

Karlsrud K Clausen CJF and Aas PM (2005) Bearing capacity of driven piles in clay the NGI approach Frontiers in Offshore Geotechnics (Proc ISFOG Perth) London Taylor amp Francis

Karlsrud K Lunne T Kort DA and Strandvik S (2001) CPTU correlations for clays Proc 16th ICSMGE 2 683-702

Kimura T Kusakabe O and Saitoh K (1985) Geotechnical model tests of bearing capacity problems in a centrifuge Geotechnique 35 (1) 33-45

Knappett JA and Craig RF (2012) Craigs Soil Mechanics 8th Edition London Spon Press Taylor amp Francis

Kolk HJ and Velde EVD (1996) A reliable method to determine friction capacity of piles driven into clays Paper No 7993 Proc Offshore Technology Conference Houston

Kolk HJ Baaijens AE and Senders M (2005) Design criteria for pipe piles in silica sands Frontiers in Offshore Geotechnics (Proc ISFOG Perth) London Taylor amp Francis Group

Kulhawy FH (2004) On the axial behavior of drilled shafts Geosupport 2004 1-18

Kulhawy FH and Jackson CS (1989) Foundation Engineering Current Principles amp Practices 2 (GSP 22) Reston VA ASCE

Kulhawy FH and Mayne PW (1990) Manual for Estimating Soil Properties for Foundation Design EPRI Report EL-6800 Electric Power Research Institute Palo Alto 306 page Download from wwwepricom

Kulhawy FH Trautmann CH Beech JF ORourke TD and McGuire W (1983) Transmission Line Structure Foundations for Uplift-Compression Loading Report EL-2870 Palo Alto CA Electric Power Research Institute wwwepricom

Ku T and Mayne PW (2015) In-situ lateral stress coefficient (K0) from shear wave velocity measurements in soils Journal of Geotechnical amp Geoenvironmental Engineering 141 (12) 101061(ASCE) GT1943-56060001354

Ladd CC and DeGroot DJ (2003) Recommended practice for soft ground site characterization Soil amp Rock America 2003 3-57

Larsson R (1997) Investigations and Load Tests in Silty Soils SGI Report R-54 Linkoumlping Swedish Geotechnical Institute

Larsson R (2001) Investigations and Load Tests in Clay Till SGI Report R-59 Linkoumlping Swedish Geotechnical Institute

Lee J and Salgado R (2005) Estimation of bearing capacity of circular footings on sands based on cone penetration tests Journal of Geotechnical amp Geoenvironmental Engineering 131 (4) 442-452

90

Lehane BM (2003) Vertically loaded shallow foundation on soft clayey silt Geotechnical Engineering 156 (1) Thomas Telford London 17-26

Lehane BM (2013) Relating foundation capacity in sands to CPT qc Geotechnical amp Geophysical Site Characterization 1 London Taylor amp Francis Group

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Lehane BM Li Y and Williams R (2013) Shaft capacity of displacement piles in clay using the CPT Journal of Geotechnical amp Geoenvironmental Engineering 139 (2) 253-266

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Low HE Lunne T Andersen KH Sjursen MA Li X and Randolph MF 2010 Estimation of intact and remoulded undrained shear strengths from penetration tests in soft clays Geotechnique 60 (11) 843-859

Lunne T Robertson PK and Powell JJM (1997) Cone Penetration Testing in Geotechnical Practice London RoutledgeBlackie Academic

Lutenegger AJ and DeGroot DJ (1995) Settlement of Shallow Foundations on Granular Soils Report Contract 6332 Task Order 4 submitted to Massachusetts Highway Dept by University of Mass-Amherst Amherst MA

Lutenegger AJ and Adams MT (2003) Characteristic load-settlement behavior of shallow foundations Proc FONDSUP 2 381-392

Mase T and Hashiguchi K (2009) Numerical analysis of footing settlement problem by subloading surface model Soils amp Foundations 49 (2) 207-220

Mayne PW (1987) Determining preconsolidation stress and penetration pore pressures from DMT contact pressures Geotechnical Testing Journal 10(3) 146-150

Mayne PW (1991) Determination of OCR in Clays by Piezocone Tests Using Cavity Expansion and Critical State Concepts Soils and Foundations 31 (2) 65-76

Mayne PW (2007) NCHRP Synthesis 368 on Cone Penetration Test Washington DC Transportation Research Board National Academies Press wwwtrborg

Mayne PW (2009) Geoengineering Design Using the Cone Penetration Test Richmond BC Canada ConeTec Inc

Mayne PW (2014) Keynote Interpretation of geotechnical parameters from seismic piezocone tests Proceedings 3rd Intl Symp on Cone Penetration Testing 47-73

Mayne PW (2016) Evaluating effective stress parameters and undrained shear strengths of soft-firm clays from CPT and DMT Australian Geomechanics Journal 51 (4) 27-55

91

Mayne PW (2017) Stress history of soils from cone penetration tests 34th Manual Rocha Lecture Soils amp Rocks Vol 40 (3) Saouml Paulo ABMS

Mayne PW Christopher B Berg R and DeJong J (2002) Subsurface Investigations -Geotechnical Site Characterization Publication No FHWA-NHI-01-031 Washington DC National Highway Institute Federal Highway Administration

Mayne PW Coop MR Springman S Huang A-B and Zornberg J (2009) Geomaterial Behavior and Testing Proc 17th Intl Conf Soil Mechanics amp Geotechnical Engineering 4 Rotterdam MillpressIOS Press

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Mayne PW and Niazi F (2017) Recent developments and applications in field investigations for deep foundations Proceedings 3rd Bolivian Conference on Deep Foundations 1 141-160

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Mayne PW and Poulos HG (1999) Approximate displacement influence factors for elastic shallow foundation systems ASCE Journal of Geotechnical amp Geoenvironmental Engineering 125 (6) 453-460

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Mayne PW and Woeller DJ (2014) Direct CPT method for footings on sands silts and clays Geocharacterization for Modeling and Sustainability (GSP 234) 1983-1997

McClelland B (1974) Design of Deep Penetration Piles for Ocean Structures Journal Geot Eng Div 100(GT7) 705-747

Mesri G (1975) Discussion on new design procedure for stability of soft clays ASCE Journal of the Geotechnical Engineering Division 101(GT4) pp 409-412

Meyerhof GG (1956) Penetration tests and bearing capacity of cohesionless soils Journal of Soil Mechanics amp Foundations Division 82 (SM1) 1-19

Meyerhof GG (1974) General Report Penetration testing outside Europe Proceedings ESOPT-1 Vol 21 Swedish Geotechnical Society Stockholm

Meyerhof GG (1976) Bearing capacity and settlement of pile foundations Journal of Geotechnical Engineering Division 102(3) 197-228

Missiaen T Verhegge J Heirman K and Crobe P (2015) Potential of cone penetrating testing for mapping deeply buried palaeolandscapes in the context of archaeological surveys in polder areas Journal of Archaeological Science 55 174-187

92

Mlynarek Z Wierzbicki J Goglik S and Bogucki M (2014) Shear strength and deformation parameters of peat and gyttja from CPTu DMT and VST Proceedings 5th International Workshop on CPTu and DMT in soft clays and organic soils 193-210 Poznan Poland Polish Committee on Geotechnics Menard Polska Tensar Internationa

Nejaim PF Jannuzzi GMF and Danziger FAB (2016) Soil behavior type of the Sarapuiacute II test site Geotechnical amp Geophysical Site Characterization 5 1009-1014

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Niazi FS and Mayne PW (2016) CPTu-based enhanced UniCone method for pile capacity Engineering Geology 212 21-34

Nowacki F Karlsrud K and Sparrevik P (1996) Comparison of recent tests on OC clay and implications for design Large Scale Pile Tests in Clay London Thomas Telford

ONeill MW (2001) Side resistance in piles and drilled shafts Journal of Geotechnical amp Geoenvironmental Engineering 127 (1) 3-16

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Pestana JM Hunt CE and Bray JD (2002) Soil deformation and excess pore pressure filed around a closed-ended pile Journal of Geotechnical amp Geoenvironmental Engineering 128 (1) 1-12

Poulos HG and Davis EH (1974) Elastic Solutions for Soil and Rock Mechanics New York Wiley amp Sons

Poulos HG and Davis E (1980) Pile Foundation Analysis amp Design New York John Wiley amp Sons

Powell JJM and Lunne T (2005) Use of CPTU data in clays and fine-grained soils Studia Geotechnica et Mechanica XXVII(3) 29-66

Randolph MF (2003) Rankine Lecture Science and empiricism in pile foundation design Geotechnique 53 (10) 847-875

Robertson PK (1990 1991) Soil classification using the cone penetration test Canadian Geotechnical Journal 27 (1) 151-158 Closure 28 (1) 176-178

Robertson PK (1991) Estimation of foundation settlements in sand from CPT Geotechnical Engineering Congress 2 Reston Virginia ASCE

93

Robertson PK (2009) Cone penetration testing a unified approach Canadian Geotechnical J 46 (11) 1337-1355

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Robertson PK and Cabal KL (2015) Guide to Cone Penetration Testing for Geotechnical Engineering 6th Edition Signal Hill California Gregg Drilling amp Testing Inc

Schmertmann JH (1970) Static cone to compute static settlement over sand Journal of the Soil Mechanics and Foundations Division 95 (SM3) 1011-1043

Schmertmann JH (1978) Guidelines for cone penetration test performance and design Report FHWA-TS-78-209 Washington DC Federal Highway Administration

Schnaid F Wood WR Smith AKC and Jubb P (1993) Investigation of bearing capacity and settlements of soft clay deposits at Shellhaven Predictive Soil Mechanics London Thomas Telford

Schneider JA Xu X and Lehane BM (2008) Database assessment of CPT-based design methods for axial capacity of driven piles in siliceous sands J Geotechnical and Geoenvironmental Engineering 134 (9) 1227-1244

Semple D (1980) Recent Developments in the Design amp Construction of Piles London Institution of Civil Engineers

Senneset K Sandven R amp Janbu N (1989) Evaluation of soil parameters from piezocone tests In-Situ Testing of Soil Properties for Transportation Transportation Research Record 1235 24-37

Shahri AA Melehmir A and Juhlin C (2015) Soil classification analysis based on piezocone penetration test data Engineering Geology 189 32-47

Stuedlein AW and Holtz RD (2010) Undrained displacement behavior of spread footings in clay The Art of Foundation Engineering Practice GSP 198 Reston Virginia ASCE

Takesue K Sasao H and Matsumoto T (1998) Correlation between ultimate pile skin friction and CPT data Geotechnical Site Characterization 2 1177ndash1182

Tand KE Funegard EG and Briaud J-L (1986) Bearing capacity of footings on clay CPT method Use of In-Situ Tests in Geotechnical Engineering GSP 6 1017-1033

Tand KE Warden PE and Funegaringrd EG (1995) Predicted-measured bearing capacity of shallow footings on sand PISCPT 2 589-594

Thomas D (1968) Deep soundings test results and the settlement of spread footings on normally consolidated sands Geotechnique 18 (2) 472-488

94

Tomlinson MJ (1957) The adhesion of piles driven into clay soils Proc 4th Intl Conf on Soil Mechanics and Foundation Engineering 2 66-71

Tomlinson MJ (1986) Foundation Design amp Construction London Longman Scientific

Tomlinson MJ (1987) Pile Design and Construction Practice London Viewpoint Publication

Trofimenkov JG (1974) Penetration testing in the USSR Proceedings ESOPT-1 1 147-154

Tumay MT Hatipkarasulu Y Marx ER and Cotton B (2013) CPTPCPT-based organic material profiling Proc 18th Intl Conf on Soil Mechanics amp Geotechnical Engineering Paris 633-636

Uzielli M and Mayne PW (2011) Statistical characterization and stochastic simulation of load-displacement behavior of shallow footings Proc Georisk 2011 Geotechnical Risk Management amp Assessment (GSP 224) 672-679

Uzielli M and Mayne PW (2012) Load-displacement uncertainty of vertically-loaded shallow footings on sands and effects on probabilistic settlement estimation Georisk Assessment and Management of Risk for Engineered Systems and GeoHazards 6 (1) 50-69

Valsson SM (2016) Detecting quick clay with CPTu Proceedings 17th Nordic Geotechnical Meeting Reykajavik Iceland Challenges in Nordic Geotechnics

Van Dijk BFJ and Kolk HJ (2011) CPT-based design method for axial capacity of offshore piles in clay Frontiers in Offshore Geotechnics II 555-560

Vesić AS (1975) Bearing capacity of shallow foundations Foundation Engineering Handbook New York Van Nostrand Reinhold Publishers

Vesic AS (1977) NCHRP Synthesis of Highway Practice 42 Design of Pile Foundations Washington DC Transportation Research Board

Yu F and Yang J (2012) Base capacity of open-ended steel pipe piles in sand J Geotechnical and Geoenvironmental Engineering 138 (9) 1116-1128

Zawrzykraj P Rydelek P and Bakowska A (2017) Geoengineering properties of Eemian peats from Radsymin (central Poland) in the light of static cone penetration and dilatometer tests Engineering Geology 226 290-300

95

APPENDIX A

List of Figures

Figure A1 Conventional methods versus direct CPT approach to shallow foundation response A-3

Figure A2 Direct bearing capacity relationship for embedded square and strip footings situated on clean

sands (after Schmertmann 1978) A-6

Figure A3 Direct CPT bearing factor Rk as function of footing width B and embedment depth from finite

element analyses (Tand et al 1995) A-6

Figure A4 Direct CPT bearing factor Rk as function of footing shape size-to-embedment ratio and sand

consistency (Eslaamizaad amp Robertson 1996) A-7

Figure A5 Direct CPT factors for bearing capacity of sands from FEM analysis by Lee amp Salgado (2005) in

terms of footing size relative density and base settlement A-8

Figure A6 Direct CPT bearing factor Rk as function of footing embedment to size ratio and normalized

cone tip resistance (after Eslami amp Gholami 2005) A-8

Figure A7 Direct relationship between ultimate bearing stress in clays and measured cone tip resistance

(Schmertmann 1978) A-10

Figure A8 Direct relationship between ultimate foundation bearing stress in clays and net cone tip

resistance (after Tand et al 1986) A-11

Figure A9 Measured load-displacements for each of the five large footings at TAMU (Briaud amp Gibbens

1994) sand site A-12

Figure A10 Characteristic nonlinear stress-normalized displacement curve for the five TAMU (Briaud amp

Gibbens 1994) footings A-13

Figure A11 Characteristic stress versus square root normalized displacements for TAMU footings A-15

Figure A12 Eurocode or LCPC criterion applied to determine bearing capacity of TAMU sand A-16

Figure A13 Summary of sand formation factors with corresponding CPT resistances A-19

Figure A14 Normalized foundation stresses vs square root of normalized displacement for spread

footings on sand (modified after Mayne et al 2012) A-20

Figure A15 Definitions of capacity from three types of observed foundation load test responses

(modified after Kulhawy 2004) A-21

Figure A16 Summary of applied footing stresses normalized to CPT net cone resistance vs pseudo-

strains (sB) for all footings A-22

A-1

Figure A17 Measured load vs displacement curves for 3 shallow foundations on clay at Baytown Texas

(Stuedlein amp Holtz 2010) A-25

Figure A18 Characteristic stress vs square root of normalized displacement for all three foundations at

Baytown site (Stuedlein amp Holtz 2010) A-26

Figure A19 Normalized foundation stress vs square root of normalized displacements A-27

Figure A20 Applied foundation stress vs CPT-calculated stress for 67 footings A-28

Figure A21 Applied footing stress vs CPT calculated stress on logarithmic scales A-29

Figure A22 Summary graph and equations of direct CPT method for evaluating the vertical A-30

Figure A23 Foundation soil formation parameter hs versus CPT material index Ic A-32

Figure A24 Bearing capacity ratio qmaxqnet versus CPT material index Ic A-33

Figure A25 Influence factor for rectangular foundations from elastic theory solution (Mayne amp

Dasenbrock 2017) A-34

List of Tables

Table A1 Direct CPT methods for bearing capacity of footings on clean sands A-4

Table A2 Direct CPT methods for bearing capacity of footings on clays A-9

Table A3 Summary of large footings soil conditions and reference sources of database A-17

Table A4 Sources for shallow foundation performace A-34

A-2

A1 INTERPRETATION OF SOIL PARAMETERS FROM CONE

PENETRATION TESTS

A11 Introduction

Geotechnical site characterization is an important first step towards the evaluation of

subsurface conditions and determination of soil layering geomaterial classification and the

evaluation of soil engineering parameters for the analysis and design of foundations retaining

walls tunnels excavations embankments and slope stability Increasing use of the electronic

cone penetrometer in highway applications is prominent in the USA because the results are

obtained much faster and less expensive than traditional methods that rely on rotary drilling

augering sampling and laboratory testing Moreover the cone penetration test (CPT) provides

at least three independent and continuous readings on soil behavior that are digitally recorded

and fully available at the end of the sounding As such the reliable and consistent

interpretation of CPT data is important since civil engineering works will best realize the

efficiency and economy of this technology in constructed facilities for county state and

interstate projects

A111 Cone Penetration Testing

The cone penetrometer is an electronically-instrumented steel probe that is vertically pushed into the

ground at constant rate of 20 mms (08 insec) using a hydraulic system The penetrometer is outfitted

with load cells pressure transducers and inclinometers that measure at least three readings

continuously with depth (a) cone tip resistance qt (b) sleeve friction fs and (c) porewater pressure u2

With the latter reading the device is called a piezocone thus the designation CPTu is used to indicate

porewater pressure Additional sensors are available that can be used to measure inclination

resistivity pH temperature shear wave velocity and other readings if desired

Figure A1 shows a schematic rendering of the cone penetration test that is performed in the

field per ASTM D 5778 procedures The hydraulic system is nominally rated at 180 kN (20 tons)

capacity and often mounted on a truck but can also be positioned on tracked vehicles or

anchored frames Figure A2 shows a MnDOT cone truck that uses the full dead weight of the

rig for the hydraulic reaction forces which are applied at the center of the vehicle The cab is

enclosed so that soundings may proceed during inclement weather and the data acquisition

system and operator are protected

Figure A 1 Setup and procedures for co ne penetration testing (CPT)

A-4

Figure A2 CPT truck-mounted rig used by MnDOT

A representative sounding taken at the I-35 test site just northeast of Saint Paul is presented in Figure

A3 Here the separate profiles of qt fs and u2 with depth are shown in side-by-side plots In the last

graph the hydrostatic porewater pressure (uo) due to the groundwater table is indicated by the blue

dashed line

These readings are used to interpret the soil layers types and properties of the ground as

discussed in the following sections

A-5

Figure A3 Representative CPT sounding at I-35 test site northeast of St Paul MN showing profiles of

(a) cone resistance (b) sleeve friction and (c) penetration porewater pressures

A112 Soil Parameter Interpretation

A variety of soil engineering parameters can be interpreted from CPT results on the basis of

theoretical frameworks analytical models or numerical simulations otherwise by empirical

methods based on correlations and statistics of databases Figure A4 shows a conceptual

utilization of the readings from the CPT for interpretation of geostratigraphy compressibility

flow characteristics and stress-strain-strength behavior of soils

Table A1 lists a selection of geoparameters that have been addressed for these purposes The

most common or useful relationships will be discussed in subsequent sections and reference is

made to other sources (Lunne et al 1997 Mayne 2007 Robertson amp Cabal 2015) for additional

details

A-6

Figure A4 Conceptual post-processing of CPT results for geoparameter evaluation

Table A1 List of common geoparameters interpreted from CPT data

Symbol Parameter Remarks Notes



γt Unit weight

ρt Mass Density

σvo Overburden stress

u0 Hydrostatic pressure

A-7

σvo Effective stress

Table A1 Continued


σp Preconsolidation stress σp = 033 (qnet)m where m = fctn (Ic)

(or Effective Yield Stress) where stresses are in kPa

YSR Yield stress ratio YSR = σpσvo (formerly OCR)

c Effective cohesion intercept

ϕ Effective friction angle


D Constrained modulus

G0 Small-strain modulus

G Shear modulus

E Youngs modulus

cvh Coefficient of consolidation

K Bulk modulus

A-8

LI

K0 Lateral stress coefficient

k Hydraulic conductivity

Liquefaction index

A12 Geostratigraphy and Soil Behavioral Type (SBT)

In routine CPT soil samples are not normally taken and thus the measured stress friction andor

porewater pressure readings are used to infer the types of soils This can be accomplished using

three basic approaches

a Rules of Thumb or approximate guidelines for a quick visual assessment

b Soil Behavioral Type (SBT) Charts that are based on either the three readings (qt fs u2) net readings

including net cone resistance (qnet = (qt-σvo) effective cone resistance (qE = qt - u2) friction ratio (Rf =

100middotfsqt) and excess porewater pressures or using normalized CPT readings such as Q F Bq or U

c Probabilistic methods where the CPT readings have been calibrated from lab tests on

recovered soil samples (eg Tumay et al 2013)

A121 Approximate Rules of Thumb

The approximate rules of thumb provide simple guidelines for soil type and suggest that sands are

identified when qt gt 5 MPa and u2 asymp u0 whereas intact clays are prevalent when qt lt 5 MPa and u2 gt u0

(Mayne et al 2002) The magnitude of porewater pressures reflects the consistency of the intact clay

such that soft (u2 asymp 2middotu0) firm (u2 asymp 4middotu0) stiff (u2 asymp 8middotu0) and hard (u2 asymp 20middotu0) However for fissured

overconsolidated clays the measured porewater pressures are less than hydrostatic in fact often

negative u2 lt 0 On land negative porewater pressure readings at the shoulder filter element of the

piezocone should be greater than -1 bar ie u2 gt -100 kPa (-147 psi) Also for clean sands the friction

ratio (FR = 100middotfsqt lt 1) while for insensitive clays (FR gt 4)

A-9

Figure A5 presents a 36-m deep piezocone sounding from the MnDOT Wakota Bridge site which crosses

the Mississippi River at I-494 (Dasenbrock 2006) The qt and u2 readings have been annotated using the

aforementioned rules of thumb indicating the predominance of sands at this site As noted there are

five interbedded clay layers evident at depths of 1 m 3-4 m 7 - 12 m 17 m and 22-27 m

Figure A5 Interpretation of soil types from CPT data taken from the Wakota Bridge on I-494 using

approximate rules of thumb

A122 Soil Behavioral Type Charts

The most common approach for identification of soil types is based on soil behavioral charts

Reviews of several chart methods are provided elsewhere (Kulhawy amp Mayne 1990 Fellenius amp

Eslami 2000 Shahri et al 2015) Current popularity favors the 9-zone classification system to

evaluate soil behavioral type (SBT) that uses normalized piezocone readings (Robertson 1990

1991 Lunne et al 1997)

A-10

(119902119905 minus 120590119907119900) (a) normalized tip resistance 119876 = A11 prime 120590119907119900

100∙119891119904 (b) normalized sleeve friction 119865() = A12

(119902119905 minus 120590119907119900)

(1199062 minus 1199060) (c) normalized porewater pressure 119861119902 = A13

(119902119905 minus 120590119907119900)

While the Q-F-Bq classification is a three-dimensional plot it is often presented as a set of two-dimensional

plots specifically Q vs F and Q vs Bq as shown in Figure A6 Data are grouped according to layers and

can be superimposed to identify their association with the nine soil behavioral types All indirect CPT soil

classification approaches should be cross-checked and verified for a particular geologic setting and local

geotechnical conditions before routine use in practice

Figure A6 Nine-zone soil behavioral type charts (a) normalized cone resistance vs normalized

sleeve friction (b) normalized cone resistance vs normalized porewater pressure parameters

(adapted after Robertson 1991 Lunne et al 1997)

The development of a CPT material index (Ic) has been found advantageous in the initial screening of soil

types and helps to organize the sounding into their respective SBT zones of similar soil response In this

case the CPT index is found from (Robertson 2009)

A-11

A2 22 )log221()log473( rtnc FQI

where Qtn = modfied stress-normalized net cone resistance given by

(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899

where σatm = reference stress equal to atmospheric pressure (1 atm asymp 1 bar = 100 kPa asymp 1 tsf asymp 147 psi)

The exponent n is soil-type dependent n = 1 (clays) n asymp 075 (silts) and n asymp 05 (sands) If units of bars are

used then Qtn (units of bars) is simply

(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900

The operational value of exponent n is found by iteration Initially an exponent n = 1 is used to calculate

the starting value of Ic (ie Qtn = Q) and then the exponent is upgraded to

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898

Then the index Ic is recalculated Iteration converges quickly and generally only 3 cycles are needed to

secure the operational Ic at each depth The soil zones and associated Ic values are detailed in Figure A7

The sensitive soils of zone 1 can be screened using the following expression

A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5

If any soil layers are found within zone 1 (sensitive soils) then caution should be exercised as these

structured clays are prone to instability collapses and difficulties in construction and performance

Another indicator of zone 1 sensitive soils is when the normalized porewater parameter Bq gt 08 When

these criteria are evident the geoengineer should consult with senior geotechnical personnel or the chief

engineer for guidance

Figure A7 Delineation of soil behavioral type zones using CPT material index Ic

A-13

Stiff overconsolidated clayey sands and sandy clays soils of zone 8 (15 lt Fr lt 45) and zone 9 (Fr gt

45) can be identified from the following criterion

Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6

The remaining soil types are identified by the CPT material index Zone 2 (organic clayey soils Ic ge 360)

Zone 3 (clays to silty clays 295 le Ic lt 360) Zone 4 (silt mixtures 260 le Ic lt 295) Zone 5 (sand mixtures

205 le Ic lt 260) Zone 6 (clean sands 131 le Ic lt 205) and Zone 7 (gravelly to dense sands Ic le 131)

The red dashed line at Ic = 260 represents an approximate boundary separating drained (Ic lt 260) from

undrained behavior (Ic gt 260)

Once the specific zone has been assigned to a layer a visual representation can be made to show

either the zone number or colorization so that the predominant layers and soil types can be

realized

A123 Probabilistic Soil Types from CPT

The inference of soil type has been addressed using probabilistic methods as discussed by Abu-

Farshakh et al (2008) and Tumay et al (2013) Here the CPT readings can be post-processed to

provide the percentage of sand silt and clay components with depth The output is consistent

and compatible with the current MnDOT soil classification chart (Figure A8) which is similar in

content to that used by the USDA

A-14

Figure A8 Chart for soil classification system by MnDOT

A13 Identifying Soft Organic Soils

The identification of soil type using cone penetration tests (CPT) is usually done on the basis of empirical

soil behavioral type (SBT) charts that use the measured readings (qt fs andor u2) While there at least

25 different sets of charts available (ie Hegazy 1998 Eslami et al 2017 Valsson 2016) one of the most

widely used and popular is the 9-zone SBT system that uses three normalized cone readings Qtn Fr and

Bq (Robertson amp Cabal 2007 Lunne et al 1997) The system is denoted as SBTn and plotted on charts in

terms of Q vs F and Q vs Bq as shown in Figure A9

A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q

Normalized Friction Fr = 100 fst(qt - svo) ()

9-Zone Soil Behavioral Type Chart

Gravelly Sands(zone 7)

Sands

(zone 6)

Sandy Mixtures(zone 5)

Silt Mix(zone 4)

Clays

(zone 3)

Organic Soils(zone 2)

Sensitive Clays

and Silts(zone 1)

Very stiff OC clayto silt (zone 9)

Very stiffOC sandto clayey

sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q

Normalized Porewater Bq = Du2(qt-svo)

Clays(zone 3)

Sensitive(zone 1)Organic

(zone 2)


Sands(zone 6)

Figure A9 CPT charts for SBTn system (a) Q versus F (b) Q versus Bq

Of particular concern in geotechnical site characterization is the proper identification of soft

organic soils such as organic clays and silts (OL and OH) and peats (Pt) These soils often cause

difficulties and problems during construction and performance of civil engineering works because

of bio-degradation long-term creep excessive settlements andor other issues

The SBTn system has a category labelled Zone 2 recognizes organic soils However several recent

studies have noted that data from CPTu soundings do not always fall within the bounds

delineated using Zone 2 even though laboratory tests and field inspections clearly note the

presence of organic soils Lab tests to classify organic soils includes loss on ignition (LOI gt 5)

and two pairs of Atterberg limits testings per ASTM standards as well as the visual-manual

identification of strong organic odor and smell and dark color notably black and dark gray

Results by Mlynarek et al (2014) show that organic soils in Poland are not adequately identified by the

Q-F chart as seen in Figure A10 Moreover studies by Zawrzykraj et al (2017) found that neither the Q-

A-16

F and Q-Bq plots worked well to recognize organic soils and peats in Poland Nejaim et al (2016) found

that Brazilian soft organic clays were misclassified as Zone 3 (clays) rather than Zone 2 (organic clays)

Similarly a recent study on CPTu data from organic clays located at 24 sites in the USA Sweden Mexico

Brazil and Australia have been compiled (Agaiby 2018) These data are shown to generally avoid falling

with the Zone 2 boundaries in either Q-F or Q-Bq charts thus are not recognized during these soundings

as indicated by Figure A11 The exception in this case is the Mexico City clay which is correctly identified

however the other 23 sites are generally not properly recognized and categorized Additional findings by

Missiaen et al (2015) found that the CPTu results in Belgian soils also did not identify organic clays and

peats in the non-normalized version of these charts that uses Qt versus friction ratio (Rf = 100middotfsqt) as

detailed by Robertson amp Cabal (2007)

Figure A10 Soil behavioral chart with superimposed CPTu data from Polish organic soils

(from Mlynarek et al 2014)

A-17

Figure A11 Paired SBTn charts with CPTu data from 24 organic clay sites indicating non-compliance

with Zone 2 (organic soils)

(compiled by Agaiby 2018)

A14 Simplified Yield Stress Evaluation in Clays by CPTu

From the development of a hybrid formulation of spherical cavity expansion (SCE) theory with critical-

state soil mechanics (CSSM) the effective preconsolidation stress or yield stress (σp) of clays can be

expressed in terms of three piezocone parameters (a) net cone tip resistance qnet = qt - σvo (b)

measured excess porewater pressure Δu2 = u2 - u0 and (c) effective cone resistance qE = qt - u2 Details

on the SCE-CSSM solutions are given elsewhere (Mayne 1991 Mayne 2017 Agaiby amp Mayne 2018)

A-18

For normal and well-behaved clays which include inorganic fine-grained soils such as clays and silts

of low sensitivity a set of simple relationships can be derived By adopting characteristic default values

for effective friction angle ϕ = 30deg and rigidity index (IR = 100) linear expressions for the effective

preconsolidation stress are obtained

prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9

A141 Application to soft Chicago clay

An example of a common geomaterial in this category includes the infamous soft Chicago clays

that were deposited in a glacio-lacustrine environment (Cho amp Finno 2010) The soft clay has a

mean water content of 20 liquid limit of 38 and plasticity index PI= 12 Results of CPTu tests

conducted at the national geotechnical experimentation site (NGES) at Northwestern University

are shown in Figure A12 (Ouyang amp Mayne 2017) As seen in the profile a thin soft clay resides

between depths of 9 to 105 m with a thicker soft clay layer in the range of 12 to 18 m

A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)

CPTu qt and u2 (kPa)

sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study

Northwestern UnivNational Test Site

Figure A12 Results of CPTu sounding at the NGES at Northwestern University Illinois

Post-processing of the CPTu data using Equations A7 A8 and A9 show good agreement in evaluating the

profile of effective yield stress in this clay layer compared with results from one-dimensional

consolidation tests made on undisturbed samples taken at the site

A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m

)Effective Yield Stress sp (kPa)

Consols

033 qnet

053 Du2

060 qE

svo

Figure A13 Evaluation of effective yield stresses at NWU using consolidation tests and CPTu

A142 Application to soft Bay Mud California

Another example of a well-behaved fine-grained soil is the San Francisco Bay Mud which is a soft clay

deposit in northern California Results of a representative CPTu are shown in Figure A14 and indicate the

soil profile at a site in eastern San Francisco (Hunt et al 2002) For this site index values include 70 lt

wn lt 90 60 lt LL lt 80 and 35 lt PI lt 45 Post-processing the CPTu data using Equations A7 A8 and

A9 show good agreement with each other as well as with the results of consolidation tests as seen in

A-21

Figure A15 The consolidation tests were performed using a CRS device as reported by Pestana et al

(2002)

San Francisco Bay Mud (Pestana et al 2002)

0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)

Cone Resistance qt (kPa)

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60

Sleeve Friction fs (kPa)

0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Figure A14 Representative CPTu in San Francisco Bay Mud (data from Hunt et al 2002)

A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)

Preconsolidation Stress sp (kPa)

svo

033 qnet

053 Du2

060 qE

CRS

svo

Figure A15 Evaluation of effective yield stresses in Bay Mud from consolidation tests and CPTu

A15 CPTu Screening of Soft Organic Clays

For soft organic clays the Equations A7 A8 and A8 will not apply thus can serve as a means to

screen such geomaterials from the soil profile Two examples of CPTu soundings in soft organic

clays are presented to illustrate the approach Additional case records and documentation of

CPTu data in organic clays are given in Agaiby (2018)

A151 Soft organic alluvial soils in Washington DC

Results of in-situ tests including CPTu soundings in soft organic clayey silts along the Potomac River at the

Anacostia Naval Air Station are presented by Mayne (1987) Here the soft soils are categorized as organic

clayey silts (OH) per the Unified Soils Classification Systems (USCS) with mean values (and plus and minus

A-23

one standard deviations) of wn = 68 plusmn 16 LL = 83 plusmn 25 and PI = 37 plusmn 17 A representative

piezocone sounding is depicted in Figure A16 For the post-processing of CPTu data the use of the Q-F

and Q-Bq graphs do not find any points within the Zone 2 category for organic soils When the data are

evaluated to determine the profile of effective yield stress the Equations A7 A8 and A9 do not agree as

shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600

Pore Pressure u2 (kPa)

Figure A16 Representative CPTu in soft organic clay along the Potomac River DC

A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter

s)Soil Behaviour Type Zone

Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)

Effective Yield Stress sp (kPa)

svo

033qnet

053 Du2

06 qE

Oed

Figure A17 Post-processing of CPTu data from Anacostia-Bolling (a) SBTn zones using CPT material

index Ic (b) CPTu screening equations for yield stress

Note that the hierarchy of the results from Equations A7 A8 and A9 show that the CPTu

evaluations for preconsolidation stress in soft organic clays indicates

[A10] σp = 053 Δu2 lt 033 qnet lt 060 qE

A152 Soft organic peaty clay in Saint Paul MN

Results of CPTu tests were collected during a training exercise underneath I-35E just northeast

of Saint Paul MN in 2007 All three MnDOT CPT rigs available at the time were used to obtain

in-situ test data at the site The site is well known as having soft organic soils and samples were

obtained from three borings made at the site (Boring ID T523 T542 and T518) Boring logs

A-25

indicate the presence of highly organic silt loams with marl and spongy organic clayey silts

From 12 samples the natural water contents ranged from 30 to 191 with a mean of 112 plusmn

58

A representative piezocone sounding from the site (MnDOT CPTu ID F22Y0703C) is used for

this illustration and is presented as Figure A18 Post-processing these data using the SBTn

algorithms and CPT material index show that the soils classify primarily as Zone 3 (clays) and

Zone 4 (silty mix) thus do not identify the geomaterials properly as Zone 2 (organic soils)

0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)

Figure A18 MnDOT CPTu sounding at the I-35E test site near St Paul Minnesota

A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q



St PaulGravelly Sands(zone 7)

Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)

Figure A19 Post-processing St Paul sounding using the 9-zone SBTn classification system

Using the recommended approach Figure A20 shows the CPT-evaluated yield stress profiles from

equations A7 A8 and A9 Here again the three estimates of σp do not agree but show the same

hierarchy as detailed in Equation A10

A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2

Figure A20 CPTu screening of soft organic soils at St Paul test site

A16 CPTu Evaluations of Yield Stress in Soft Organic Soils

Once the presence of soft organic clays and silts is identified using the aforementioned CPT

screening algorithms the evaluation of effective yield stress is conducted using the procedure

outlined in Chapter 2 of this MnDOT manual For soft organic soils an exponent of m = 090 is

recommended (Mayne et al 2009) Therefore the preconsolidation stress is obtained from

(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090

The algorithm can be expressed in dimensionless form by

prime = A12 120590119901 033 ∙ (119902119905 minus 120590119907119900)119898prime ∙ (120590119886119905119898100)1minus119898prime

A-28

where σatm = reference stress (= 100 kPa = 147 psi) and m = 090 for soft organic silts and clays

The approach is applied to the two former example case studies in Figure A21 (Washington DC)

and Figure A22 (St Paul MN) respectively showing good agreement with benchmark results

taken from laboratory consolidation tests on undisturbed samples of these sites in both cases

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed

Figure A21 Profiles of yield stress from consolidation tests and CPTu method for organic clayey silts at

Anacostia-Bolling site in Washington DC

A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa


MnDOT test site near Saint Paul Minnesota

A17 Soil Unit Weight

As shown by Figure A23 total soil unit weight (t) can be estimated from the CPT sleeve friction resistance

(Mayne 2014) The equation can be expressed by

120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13

where w = unit weight of water and satm = atmospheric pressure

A-30

Figure A23 Evaluation of soil unit weight from CPT sleeve friction

A18 Effective Stress Friction Angle

Sands

The effective stress friction angle () is one of the most important soil properties as it governs the strength

of geomaterials as well as affects soil-pile interface and pile side friction While an effective cohesion

intercept (c) can also be considered this is usually reserved for cemented or bonded geomaterials or

unsaturated soils and may lose its magnitude with time ageing or with prolonged environmental

changes

For clean quartz to silica sands where porewater pressures are essentially hydrostatic (Bq = 0) the

following expression has been calibrated with triaxial compression test results from undisturbed sand

samples and normalized cone resistances as presented in Figure A24 (Mayne 2007 2014)

A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14

Figure A24 Evaluation of effective stress friction angle in quartz-silica sands from CPT

Note Relationship applies to drained soil behavior when Ic lt 26 andor Bq lt 01

where qt1 is an earlier form of stress normalized cone tip resistance for sands given by

(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32

In terms of units of bars this is expressed simply as (in units of bars)

119902119905 1199021199051 = 05 A152 prime ) (120590119907119900

Recently Robertson amp Cabal (2015) recommended the use of the modified form of normalized cone

resistance (Qtn) in Equation A10

120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16

In sands qnet asymp qt since the overburden term is small relative to the cone tip resistance Figure A25 shows

that the two normalizations give very comparable results

Figure A25 Comparable values from qt1 and Qtn normalization schemes for CPTs in sands

Clays and Silts

A-33

In the case of soft to firm intact clays and silty clays the effective friction angles are determined from the

normalized cone resistance and porewater pressure parameters (Senneset et al 1989 Mayne 2016) as

shown in Figure A26 The exact solution when the angle of plastification = 0 is given as

qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17

Approximate NTH Solution for from CPTu

qBQ

)tan1(tan61

1)tanexp()245(tan2

Approx asymp 295deg∙Bq0121∙[0256 + 0336∙Bq + log Q ]

1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)

Approximation (lines)

Theory

Figure A26 Evaluation of effective stress friction angle in clays and silts from CPTu results

Note Relationship applies to undrained soil behavior when Ic ge 26 andor Bq ge 01

which can be approximately inverted into the form (Mayne 2007)

A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18

This algorithm is specifically applicable for the following ranges of porewater pressure parameter (01 le

Bq lt 1) and effective stress friction angles (20deg le lt 45deg)

A19 Stress History

The stress history can be characterized by an apparent yield stress or preconsolidation stress (sp) as well

as by its normalized and dimensionless form YSR = spsvo = yield stress ratio The YSR is in effect the

same as overconsolidation ratio (OCR) however now generalized to accommodate mechanisms of

preconsolidation that occur beyond just erosion glaciation and removal of overburden stresses but also

due to ageing desiccation repeated cycles of wetting-drying bonding repeated freeze-thaw cycles

groundwater changes and other factors

A-35

A generalized approach for evaluating the yield stress or preconsolidation in soils using net cone

resistance has been formulated as presented in Figure A27 (Mayne 2015) Yield stress in soils from CPT

10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)

Net Cone Resistance qt - svo (kPa)

Blessington

General Trend

sy = 033(qt-svo)m

Fissured Clays m = 11

Intact clays m = 10 Sensitive clays m = 09

Silt Mixtures m = 085Silty Sands m = 080

Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays

Figure A27 Evaluation of yield stress or preconsolidation stress in soils from CPT


1minus119898prime prime 120590119886119905119898 120590119901 = 033 ∙ (119902119905 minus 120590119907119900)119898prime

( ) A19 100

where m = exponent depends on soil type m = 1 (intact clays) 085 (silts) 080 (sandy silts to silty sands)

and 072 (sands) For fissured clays the exponent m may be 11 or higher depending upon the age of the

A-36

formation and degree of jointing and discontinuities Note that fissured clays can be identified when Ic lt

26 and Bq lt 01

The exponent m has also been calibrated with CPT material index as presented in Figure A28

An algorithm to express this relationship for non-fissured soils is given by

25)652(1

2801

cIm

A20

This allows the post-processing of CPT to automatically choose the appropriate exponent m for

a layer by layer analysis

Figure A28 Yield stress exponent m in terms of CPT material index Ic

A-37

A110 Lateral Stress Coefficient

The horizontal geostatic state of stress is represented by the lateral stress coefficient K0 = shosvo

commonly referred to as the at-rest condition The magnitude of K0 for soils that have been loaded and

unloaded can be approximately estimated from

119870119900 = (1 minus 119904119894119899120601prime) ∙ 119874119862119877119904119894119899120601prime A21

Data compiled from in-situ K0 measurements using self-boring pressuremeter tests (SBPMT) total stress

cells (TSC) or push-in spade cells andor laboratory methods (instrumented consolidometers triaxials

andor suction measurements) on a variety of clays silts and sands have been compiled and reported by

Ku amp Mayne (2015) as shown in Figure A29 verifying Equation A15 as a means for evaluating K0 in soils

Figure A29 Relationship between lateral stress coefficient K0 and YSR or OCR in soils (Ku amp Mayne

2015)

A-38

A111 Undrained Shear Strength

The loading of soils can occur under conditions of being fully drained (Du = 0) partially drained or full

undrained (DVV0) where Du = excess porewater pressures (above hydrostatic) and DVV0 = volumetric

strain Further details on specific stress paths are best explained in terms of critical state soil mechanics

or CSSM (Mayne et al 2009 Holtz Kovacs amp Sheahan 2011) The prevailing drainage conditions depend

upon the rate of loading and permeability characteristics of the soil Normally in sands that are pervious

and exhibit high permeability a drained response occurs Exception to this may occur in loose sands

during fast earthquake loading resulting in soil liquefaction In clays that exhibit low permeability a fast

rate of loading will result in undrained loading at constant volume This in fact is a temporary and

transient condition often termed short term loading In the long term eventually porewater pressures

will dissipate (albeit slowly) and a drained response will prevail thus termed long term loading

The overall soil strength is controlled by the effective stress strength envelope Most commonly this is

represented by a simple linear relationship termed the Mohr-Coulomb criterion where the maximum

shear stress (max called the shear strength) is given as

= 119888prime + 120590prime ∙ 119905119886119899 120601prime A22 120591119898119886119909

where c = effective cohesion intercept s = effective normal stress and = effective stress friction angle

As a starting point values of c = 0 and = 30deg can be adopted for all soil types at least until the specific

soil type geologic formation and results from high-quality laboratory or field test data are available

Peak Undrained Shear Strength from Stress History

Using simplified critical state soil mechanics the peak undrained shear strength (max = su) can be

evaluated from the effective stress strength envelope (c = 0 effective friction angle ) and stress history

(ie OCR) in the form

prime DSS 119904119906 = (119904119894119899120601prime2) ∙ (119874119862119877)Λ ∙ 120590119907119900 A23

A-39

which corresponds to a simple shear mode Figure A30 presents the aforementioned relationship

together with data from 17 different clays tested in simple shear

Figure A30 Normalized undrained shear strength from simple shear tests on various clays

versus YSR or OCR

For the triaxial compression (TC) mode the equation would give slightly higher strengths that are

calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24

where Mc = (6 middot sin)(3-sin)

Peak Undrained Shear Strength from CPTu

A more direct approach to assessing undrained shear strength is via bearing capacity theory

whereby

A-40

A25 kt

votu

N

qs

s

where Nkt = bearing factor that depends on mode of shearing sensitivity of the clay degree of

overconsolidation and other factors For soft offshore clays the back figured Nkt from data collected at

14 well-documented sites (Low et al 2010) determined mean values based on mode of shearing Nkt =

119 (triaxial compression) Nkt = 136 (simple shear) and Nkt = 133 (vane)

A recent study of 51 clays that were tested by both field piezocone and laboratory CAUC triaxial tests

showed that essentially Nkt = 12 for intact clays and clayey silts of low to medium sensitivity (Mayne

Peuchen amp Baltoukas 2015) Figure A31 shows the slightly different trends for offshore versus onshore

clays For sensitive clays a lower value of Nkt = 10 would be appropriate and for fissured clays a higher

value of Nkt would be in the range of 20 to 30

A-41

A-42

Figure A31 Database from 51 clays with CPT qnet versus lab measured triaxial shear strength

(Mayne Peuchen amp Baltoukas 2015)

For soft to firm clays an alternate means to evaluate undrained shear strength is via the excess porewater

pressures ( u = u2 - u0) from the following

u

uN

us

D

D 2

A26

where N u = porewater bearing factor also dependent on the aforementioned factors The study by

Low et al (2010) found N u = 59 (triaxial compression) N u = 69 (simple shear) and N u = 71 (vane

shear)

A third alternate is found from the effective cone resistance (qE = qt - u2) whereby

kE

tu

N

uqs 2

A27

where NkE is a bearing term for this approach Mayne et al (2015) indicated a mean value of NkE = 8 for

soft-firm intact clays

Remolded Undrained Shear Strength from CPT

The remolded undrained shear strength (sur) is obtained from either field vane tests lab mini-vane shear

tests or lab fall cone devices This affords the evaluation of the clay sensitivity (St) which is defined as the

ratio of peak to remolded strengths at a given water content

119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903

For the CPT the sleeve friction has been noted to give values that are comparable to the

remolded undrained shear strength (Powell amp Lunne 2015) Therefore

asymp 119891119904 A29 119904119906119903

A112 Ground Stiffness and Soil Moduli

The stiffness of the ground can be represented by several geoparameters including the compressibility

indices (Cr Cc Cs) spring constants (kv) and moduli Regarding the latter there are several moduli that

are used in geotechnical engineering including shear modulus (G and Gu) Youngs modulus (E and Eu)

constrained modulus (D) bulk modulus (K) resilient modulus (MR) and subgrade reaction modulus (ks)

A-43

Soil Modulus

The definition of modulus is taken as E = DsD with Figure A32 showing a typical deviator stress (q = s1

- s3) versus axial strain (a) curve Theoretical interrelationships between the elastic moduli G E D and

K have a dependence on the Poissons ratio v The value of Poissons ratio can be taken as vu = 05 for

undrained loading (ie constant volume) while for drained loading which is accompanied by volumetric

strains a value of v = 02 may be used If we adopt the reference modulus as E then the

interrelationships with the other elastic moduli are given by

a Reference Stiffness 119864prime = drained Youngs modulus A30

119864prime

b Shear Modulus 119866prime = A31 [2(1+120584prime)]

119864prime (1minus120584prime) c Constrained Modulus 119863prime = A32

[(1+120584prime)(1minus2120584prime)]

119864prime

d Bulk Modulus 119870prime = A33 [3middot(1minus2120584prime)]

A-44

Figure A32 Representative stress-strain-strength curve for soils in triaxial compression

For the extreme case when = 0 in fact D = E When = 02 the two moduli are only 10

different D = 11middotE thus the constrained modulus and drained Youngs modulus are often

considered somewhat interchangeable

The resilient modulus (MR) applies to pavement analysis and design most commonly measured by cyclic

triaxial testing under repeated load applications In fact MR is a special case of Youngs modulus that

relates to the small strain stiffness measured in the nondestructive range but has developed permanent

plastic strains after many cycles of loading (Brown 1996 Dehler amp Labuz 2007)

The subgrade reaction modulus is actually a combined soil-structural parameter as its value

depends on the ground stiffness and the size of the loaded element The subgrade modulus is

defined as ks = q where q = applied stress and = measured deflection In terms of elasticity

solutions the deflection of a flexible circle of diameter d is given by = qmiddotdmiddot(1-2)E Therefore

119864prime

119896119904 = A34 [119889middot(1minus1205842)]

which has units of kNm3 or pcf

Table A2 summarizes several selected studies towards the approximate evaluation of these moduli

obtained directly from measured CPT data Note however that the results from in-situ penetrometer

tests generally represent a peak strength as indicated in Figure A32 Thus the measured cone tip

resistance (qt) reflects the top of the stress-strain curve either the undrained strength in clays (su) or the

effective friction angle () of sands or even a strength intermediate between these two values Thus

A-45

Source Soils Studied Reference Modulus

Findings

Kulhawy amp Mayne 1990

8 stiff OC clays Lab constrained modulus

D asymp 825 middot (qt - svo)

Mohammed et al (2000)

Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites

Lab constrained modulus


Mayne (2007b)

All soil types Constrained moduli from lab consolidation

First-order estimate D asymp middot(qt - svo)

Robertson (2009)

All soil types Constrained modulus from consolidation

D = m middot (qt - svo)

when Ic gt 22

1 m = Qtn when Qtn le 14

2 m = 14 when Qtn gt 14

when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)

16 clay sites in China

Resilient modulus MR = (146qt 053 + 1355 fs

14 + 236)244

(all in MPa)

Casey et al (2016)

8 clays Triaxial tests for undrained Youngs modulus

Eu(svc)07 = 316 - 23 middot LL

where Eu and svc (MPa)

and LL = liquid limit ()

qt is probably not the best means to obtain the slope of the stress-strain curve Instead the SCPTu that

provides the initial tangent shear modulus (Gmax) from the shear wave velocity (Vs) is a better choice

Table A2 Selected Modulus Relationships with Cone Penetration Test Measurements

A-46

Nonlinear Modulus

The small-strain shear modulus (Gmax or G0) represents the initial stiffness of all soils and rocks It is the

beginning of all stress-strain-strength curves for geomaterials and obtained from elasticity theory

119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

where Vs = shear wave velocity t = tga = total soil mass density t = total soil unit weight and ga =

gravitational acceleration constant (98 ms2)

From a general viewpoint on stiffness the shear modulus G of soil is defined as the slope of shear stress

versus shear strain G = DDs for a tangent definition and by G = s as a secant definition The shear

modulus is related to its associated Youngs modulus E = 2G(1+v) Both moduli are in fact highly

nonlinear ranging from a maximum value at the small-strain stiffness (Gmax = G0 = t middotVs2 where t =

total soil mass density and Vs = shear wave velocity) to intermediate G values at medium strains (asymp 1)

to low values at peak strength As such a variety of algorithms and formulae have been developed to

represent either a partial range or the full stress-strain-strength behavior of soils over a range of

interests (eg Mayne 2005) In these formulations a variable number of input parameters may be

required in order to produce a stress-strain-strength curve

A modified hyperbolic form suggested by Fahey amp Carter (1993) has favorable attributes in that the

modulus reduction factor can be established with only a single variable This allows the initial stiffness

to the be small-strain stiffness (G0) that is reduced in terms of level of mobilized strength eg max =

qqmax = 1FS which is simply the reciprocal of the calculated factor of safety (FS) The magnitude of

secant shear modulus (G) corresponding to the particular level of loading is given by

119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47

where the MRF = modulus reduction factor determined as

1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909

and g = empirical exponent term Figure A33 shows the relationships for normalized shear stress ()

versus shear strain (s) and corresponding normalized secant shear modulus (G = s) with shear strain

over a range of exponent values 01 le g le 10 For a discussion of tangent G please see Fahey amp Carter

(1993)

Using laboratory data from resonant column-torsional shear tests and triaxial specimens with local

strain measurements for Gmax reference values modulus reduction curves for a selection of sands and

clays are presented in Figure A34 The data indicate that the exponent g falls within the range 02 lt g lt

05 for many soils tested drained and undrained A mean value of g = 03 is recommended for

preliminary site investigations and designs until additional information can be obtained

The value of max is the shear strength of the soil generally taken as either (1) drained (c = 0) max =

svo middot tan or (2) undrained where max = su = undrained shear strength as discussed earlier Thus each

shear stress () can be associated with its shear modulus (G) and the relevant shear strain is found from

s = G This allows for generation of nonlinear stress-strain-strength curves at all depths from SCPTu

data in clays sands and mixed soil types

A-48

Modified Hyperbola (Fahey amp Carter 1993 Canadian Geotech J) Coefficient Term f = 1

``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x

Mobilized Shear Stress max

g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g

Figure A33 Normalized modulus (GGmax) and normalized shear stress (max) versus shear strain (s)

for modified hyperbolic relationship of Fahey amp Carter (1993)

A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x

Mobilized Strength max or qqmax

NC SLB Sand

OC SLB Sand

Hamaoka Sand (e=0625)

Hamaoka Sand

Toyoura Sand e = 067


Ham River Sand

Ticino Sand

Kentucky Clayey Sand

Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g

Figure A34 Modulus Reduction Factors (MRF) with mobilized strength of clays and sands

A113 Coefficient of Consolidation

The rate at which foundation and embankment settlements occur as well as the dissipation of excess

porewater pressures is controlled by the coefficient of consolidation (cv) The magnitude of cv is also

required in the design of vertical wick drains that can be installed in soft ground to expedite the time for

consolidation Using the results of CPTu dissipation tests which measure the rate at which the u2 readings

vary with time the in-situ profile of cv can be evaluated Often the piezocone-interpreted values of cv

are validated by comparison with results from laboratory one-dimensional consolidation tests (eg Abu-

Farsakh MY 2004) Alternatively a better method is to cross-check the values with the measured full-

scale performance of instrumented embankments that are constructed over the ground and the recorded

time-rate of consolidation and settlements can provide the best cv for that site (Abu-Farsakh MY et al

2011)

A-50

Monotonic Dissipation Tests

An illustrative dissipation curve from piezocone testing at IDT State Highway 95 at an embankment and

bridge crossing is presented in Figure A35 After the CPTu sounding was halted at a depth of 512 feet

the recorded decay of porewater pressures was observed to be monotonic with time until the dissipations

were ended at 1000 seconds During that time the u2 readings decreased from 115 psi to 47 psi At this

location the depth to the groundwater table is about 16 feet thus the calculated equilibrium water

pressure is 15 psi Commonly a characteristic time for piezo-dissipations is taken at 50 degree of

consolidation although other degrees may be adopted By evaluating the value of u2 at 50 as shown in

Figure A35 the characteristic t50 = 404 s can be obtained

0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)

u2 (initial) = 115 psi

u2 (final projected) = u0 = 15 psi

u2 at 50 = frac12(115+15) = 65 psi

Idaho DOT State Route 95Dissipation at 512 feet

t50 = 404 seconds

Time to reach50 consolidation

A-51

Figure A35 Piezo-dissipation at Sandpoint Bridge Crossing Idaho for depth z = 512 feet illustrating the

evaluation of characteristic time t50

It is also common to plot normalized excess porewater pressures relative to the measured initial Du2

that is obtained during penetration at the constant rate of 20 mms ie DuDui versus time Figure A36

shows a set of piezo-dissipation records for a bridge and embankment site in southern Louisiana

reported by the LTRC While the porewater pressure axis is arithmetic the time scale is often plotted on

either logarithmic or square root scales In either case the t50 is then found from the measured

dissipation curve when DuDui = 050

A-52

Figure A36 Set of monotonic piezo-dissipations taken at various depths for Courtableau Bridge

Site on Louisana State Highway 103 (Abu-Farsakh et al 2011)

Dilatory Dissipation Curves

In a number of situations a monotonic type dissipation does not occur on the onset but instead a dilatory

type curve is observed Here after stopping the push of the penetrometer the recorded porewater

pressures initially begin to rise and eventually reach a peak value then afterwards follow a phase where

the Du values decrease to hydrostatic (Figure A37) A full solution is available for both cases (Burns amp

Mayne 2002) Dilatory responses are most often associated with overconsolidated clays and silts

although occasionally are observed in fine-grained soils with low OCRs

The selection of the characteristic t50 value may found by using a square root of time plot to find the initial

u2 value by projection of the post-peak dissipatory data back to the ordinate axis as shown by the example

in Figure A38 In this example the initial u2 = 480 kPa and then the same procedures in Figure A35may

apply

A-53

Figure A37 Monotonic and dilatory porewater dissipation responses in soils

(after Burns amp Mayne 1998)

Figure A38 Method for obtaining t50 from dilatory dissipation curves

(Schneider amp Hotstream 2010)

A-54

Interpretation of Dissipation Test Data

There are a number of different solutions available for the interpretation of the coefficient of

consolidation (cv) from the piezocone dissipation curves For monotonic type response the strain path

method (SPM) developed at Oxford University (Teh amp Houlsby 1991) is well-recognized while the Georgia

Tech solution by Burns amp Mayne (2002) is based on spherical cavity expansion and critical state soil

mechanics (SCE-CSSM) and handles both monotonic and dilatory curves For both solutions a simplified

procedure can be recommended that relies on the aforementioned t50 value obtained from the measured

field data as given in Table A3 The units for cv are in cm2s as shown or equivalent

Table A3 Recommended procedures for calculating cv from t50 obtained in dissipation tests

Method Equation for coefficient of

consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v

t50 = measured time to reach 50

dissipation

IR = Gsu = undrained rigidity index

G = shear modulus

su = undrained shear strength

ac = penetrometer radius (ac = 178

cm for 10-cm2 cone ac = 220 for a

15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55

Evaluating rigidity index

Both the SPM and SCE-CSSM solutions require an estimate of the in-situ rigidity index (IR = Gsu) of the

soil If the results of SCPTU are available Krage et al (2014) have derived an expression for IR that

depends upon the small-strain shear modulus net cone tip resistance and effective overburden stress

250750

max50

8111

vonet

Rq

GI

s A38

where consistent units are input for Gmax qnet and svo The value of Gmax is obtained from B29 using the

measured shear wave velocity (Vs) and unit weight evaluated from B7

If the shear wave velocity is not measured it can be estimated from the CPT data A number of

methods have been reviewed for CALTRANS by Wair et al (2012) including one that is applicable

to sands silts and clays of low sensitivity that are inorganic and uncemented

119881 (119898119904) = [101 middot 119897119900119892(119902119905) minus 114]167 middot (100 middot 119891119904119902119905)03 A39 119904

where qt is in units of kPa (Hegazy amp Mayne 1995) An alternative relationship is given by Robertson amp

Cabal (2015)

A114 Hydraulic Conductivity

The hydraulic conductivity is a parameter that expresses the flow characteristics of the soil In

geotechnics it is also called the coefficient of permeability (k) and has units of cms or feetday

Through consolidation theory the hydraulic conductivity relates directly to the coefficient of

consolidation

A-56

A40 D

ck wv

where w = unit weight of water and D = constrained modulus

Therefore one approach to evaluating k can be from the site-specific cv obtained from the

dissipation tests using an estimate of D from one of the relationships given in Table A2

An approach developed for soft normally-consolidated soils that uses t50 directly to assess the magnitude

of k is presented in

Figure A39 (Parez and Fauriel 1988) The mean trendline through the wedges of soil type can be

approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57

Figure A39 Hydraulic conductivity versus dissipation time for 50 consolidation

(Parez amp Fauriel 1988)

A-58

APPENDIX B

List of Figures

Figure B1 Conventional methods vs direct CPT approach to shallow foundation response B-3

Figure B2 Direct bearing capacity relationship for embedded square and strip footings situated on clean

sands (after Schmertmann 1978) B-6

Figure B3 Direct CPT bearing factor Rk as function of footing width B and embedment depth from finite

element analyses (Tand et al 1995) B-6

Figure B4 Direct CPT bearing factor Rk as function of footing shape size-to-embedment ratio and sand

consistency (Eslaamizaad amp Robertson 1996) B-7

Figure B5 Direct CPT factors for bearing capacity of sands from FEM analysis by Lee amp Salgado (2005) in

terms of footing size relative density and base settlement B-8

Figure B6 Direct CPT bearing factor Rk as function of footing embedment to size ratio and normalized

cone tip resistance (after Eslami amp Gholami 2005) B-8

Figure B7 Direct relationship between ultimate bearing stress in clays and measured cone tip resistance

(Schmertmann 1978) B-10

Figure B8 Direct relationship between ultimate foundation bearing stress in clays and net cone tip

resistance (after Tand et al 1986) B-11

Figure B9 Measured load-displacements for each of the five large footings at TAMU (Briaud amp Gibbens

1994) sand site B-12

Figure B10 Characteristic nonlinear stress-normalized displacement curve for the five TAMU (Briaud amp

Gibbens 1994) footings B-13

Figure B11 Characteristic stress versus square root normalized displacements for TAMU footingsB-15

Figure B12 Eurocode or LCPC criterion applied to determine bearing capacity of TAMU sandB-16

Figure B13 Summary of sand formation factors with corresponding CPT resistancesB-19

Figure B14 Normalized foundation stresses vs square root of normalized displacement for spread

footings on sand (modified after Mayne et al 2012) B-20

Figure B15 Definitions of capacity from three types of observed foundation load test responses

(modified after Kulhawy 2004) B-21

Figure B16 Summary of applied footing stresses normalized to CPT net cone resistance vs pseudo-

strains (sB) for all footingsB-22

B-1

Figure B17 Measured load vs displacement curves for 3 shallow foundations on clay at Baytown Texas

(Stuedlein amp Holtz 2010)B-25

Figure B18 Characteristic stress vs square root of normalized displacement for all three foundations at

Baytown site (Stuedlein amp Holtz 2010) B-26

Figure B19 Normalized foundation stress vs square root of normalized displacements B-27

Figure B20 Applied foundation stress vs CPT-calculated stress for 67 footingsB-28

Figure B21 Applied footing stress vs CPT calculated stress on logarithmic scales B-29

Figure B22 Summary graph and equations of direct CPT method for evaluating the vertical B-30

Figure B23 Foundation soil formation parameter hs versus CPT material index Ic B-32

Figure B24 Bearing capacity ratio qmaxqnet versus CPT material index IcB-33

Figure B25 Influence factor for rectangular foundations from elastic theory solution (Mayne amp

Dasenbrock 2017) B-34

List of Tables

Table B1 Direct CPT methods for bearing capacity of footings on clean sandsB-4

Table B2 Direct CPT methods for bearing capacity of footings on clays B-9

Table B3 Summary of large footings soil conditions and reference sources of database B-17

Table B4 Sources for shallow foundation performace B-34

B-2

B1 DIRECT CPT METHOD FOR SHALLOW FOUNDATIONS

B11 Introduction

The analysis of shallow foundations on soils is classically handled as a two-step and two-part set of

calculations involving (a) bearing capacity commonly reliant on limit plasticity solutions and (b)

settlement or more appropriately termed displacements that are assessed via elastic continuum theory

The utilization of cone penetration tests (CPT) provides the necessary geotechnical data for the site-

specific information on the subsurface conditions at the project site including the geostratigraphy and

evaluation of input parameters This is still a viable approach where the CPT readings are interpreted to

give the soil unit weight (γt) effective friction angle (ϕ) undrained shear strength (su) preconsolidation

stress (σp) and elastic moduli (D and E) for analysis

An alternative approach is the use of CPT results to provide direct assessments of bearing capacity

andor settlements A comparison of these two distinctly different and alternate paths is depicted in

Figure B1

Figure B1 Conventional methods versus direct CPT approach to shallow foundation response

A number of available direct CPT approaches for shallow footings are reviewed within this report

Moreover as the entire load-displacement-capacity response of shallow foundations to loading occurs

as a continuous nonlinear phenomenon a single direct CPT solution is presented for the behavior of

B-3

footings on sands (drained) and clays (undrained) The methods discussed herein are specifically to

address the general case of vertical loading of shallow foundations Additional more complex situations

that consider load eccentricity moments inclined forces sloping ground and other facets may be

handled using well-established procedures that are discussed elsewhere (eg Vesić 1975 Kulhawy et

al 1983)

This report focuses on foundations on granular soils andor soils exhibiting drained behavior since

Federal Highway Administration (FHWA) studies have concluded that less than 1 of shallow

foundations for highway bridges are placed on clay soils (Paikowsky et al 2010) One reason for this is

because soft-firm intact clays require a more complicated process that assesses both a short-term

analysis (undrained) as well as long-term analysis (drained) thus requiring undrained bearing capacity

undrained distortion displacements drained bearing capacity and drained primary consolidation

settlements

B12 Direct CPT Methods for Bearing Capacity of Footings on Sands

Classical bearing capacity solutions are based most often in limit plasticity theory albeit can be

formulated from limit equilibrium cavity expansion and numerical methods Because the limit

plasticity solutions are prevalent and adopt total stress analysis they are usually developed as either

fully drained (Δu = 0) or fully undrained (ΔVV = 0) where Δu equals excess porewater pressure and

ΔVV equals volumetric strain Paikowski et al (2010) provides an extensive review of various solutions

There are a number of direct CPT methods for the evaluation of the bearing capacity of footings and

shallow foundations Table B1 lists a variety of these direct CPT approaches for footings on sands In

the direct approach a one-step process is used to scale the measured cone penetrometer readings (ie

measured cone tip resistance (qc) total cone tip resistance (qt) sleeve friction (fs) andor measured

porewater pressure u2)) to obtain the ultimate bearing stress (qult) of the foundation

Table B1 Direct CPT methods for bearing capacity of footings on clean sands

Method Surface Footing Remarks and Embedment Notes

Meyerhof (1956) qult = qc (B12) middotcw qult = qc (B12)(1+DeB)cw

Note for silty sand reduce by 05

cw = 10 dry or moist sand cw = 05 submerged sand

Meyerhof (1974) qult = qc (B + D)40 with stresses in tsf

Presumably dry sands B = fdn width (feet) and D = depth (feet)

Schmertmann (1978)

NA (applies to embedded footings)

Square qult =055σatm(qcσatm)078

Strip qult =036σatm(qcσatm)078

See Figure A2

Embedment applies De gt 05(1+B) for Blt1m De gt 12 m for B gt1m

B-4

Canadian qult = Rk0middotqc Applied to FS = 3 Geotech Society where Rk0 = 03 where FS = factor of (CFEM 199 2) safety

Tand et al (1995) NA qult = Rkmiddotqc + σvo See Figure A3 where Rk = fctn(De B)

Frank and Magnan (1995)

qult = Rk0middotqc where Rk0 = fctn(sand

consistency) Loose Rk0 = 014

Medium Rk0 = 011 Dense Rk0 = 008

qult = Rk1middotqc + σvo where Rk1 = function (De B L amp

sand consistency) Factor Rk1 = Rk0[1+Rk206+04(BL) (DeB)]

and Rk2 = 035 (loose) 050 (medium) and 085 (dense)

Loose sand qc lt 5 MPa

Medium sand 8 MPa lt qc lt 15MPa

Dense sand qc gt 20 MPa

Eslaamizaad amp Robertson (1996)

qult = KΦmiddotqc See Figure A4

See surface footing equation KΦ = function (BDe shape and density)

Lee amp Salgado (2005)

qbL = βbcmiddotqc(AVG)

where qc averaged over distance B

Not addressed See Figure A5 for factor βbc = fctn(B DR

K0 and sB)

beneath the base

Eslami amp Gholami (2005

2006)

See embedded footing solution

qult = Rk1middotqc where Rk1 = function (ratio DeB

and normalized qcσvo) See Figure A6

Measured qc and qcσvo are geometric means over 2B deep

beneath footing

Robertson amp Cabal (2007)

qult = KΦmiddotqc with KΦ = 016

See surface footing equation

Briaud (2007) qult = KΦmiddotqc with KΦ = 023

Based on full-scale tests at Texas AampM

Mayne amp Illingworth

(2010) qult = 018 qc-Mean

Note qc -Mean is averaged CPT cone resistance over depth of

influence z = 15 B deep

Based on 30 footing load tests on 12

sands

Lehane (2013) qult = 016 qc-AVE Note qc-AVE is averaged CPT cone resistance within z (m) = [B (m)

]07

Based on 47 load tests

Table B1 Continued

Notes B = minimum footing width (or diameter) L = footing length De = embedment depth cw = water

table correction σvo = total overburden stress at bearing elevation and σvo = the effective vertical stress

B-5

Figure B2 Direct bearing capacity relationship for embedded square and strip footings

situated on clean sands (after Schmertmann 1978)

Figure B3 Direct CPT bearing factor Rk as function of footing width B and embedment depth

from finite element analyses (Tand et al 1995)

B-6

Figure B4 Direct CPT bearing factor Rk as function of footing shape size-to-embedment ratio

and sand consistency (Eslaamizaad amp Robertson 1996)

B-7

Figure B5 Direct CPT factors for bearing capacity of sands from FEM analysis by Lee amp

Salgado (2005) in terms of footing size relative density and base settlement

Figure B6 Direct CPT bearing factor Rk as function of footing embedment to size ratio and

normalized cone tip resistance (after Eslami amp Gholami 2005)

B13 Direct CPT Methods for Bearing Capacity of Footings on Clays

For direct CPT evaluations of foundation bearing capacity in clays Table B2 summarizes some of the

well-documented methods that are available Generally these assume that the loading is fast enough

and the permeability of the clay is also sufficiently low such that an undrained (ie constant volume)

condition is maintained This is fine for short term loading of foundations particularly for soft to firm

intact clays However the undrained case is not permanent Given sufficient time excess porewater

pressures in the clay will eventually dissipate and hydrostatic conditions will return to equilibrium

During this dissipation phase primary consolidation will occur that results in drained settlements Also

a drained bearing capacity condition will prevail As the CPT is advanced at a constant rate of 20 mms

the recorded readings normally constitute undrained behavior from a direct measurement viewpoint

Thus direct CPT methods are normally focused at an undrained foundation response Nevertheless the

drained footing case can be addressed using CPT results via use of conventional analysis approach and

effective stress limit plasticity solutions that require piezocone penetrometers with porewater pressure

measurements

B-8

B

D)

L

B4060(3501320kc

Many of the earlier solutions were based on data from mechanical penetrometers andor older electric

cones that measure the uncorrected tip resistance (qc) It is now well-recognized that this measurement

must be updated to the corrected cone resistance (qt) because of porewater pressure effects that act on

the mechanics of the load cell and geometry of the particular penetrometer design The results are

particularly significant in clays silts and mixed soils that are soft to firm to stiff where excess porewater

pressures are recorded

Table B2 Direct CPT methods for bearing capacity of footings on clays

Method Footing Comments NotesRemarks

Meyerhof (1974) qult = αbcmiddotqc

where 025 le αbc le 050

Applicable to saturated insensitive clays under short-term loading

Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)

Strip footings on clays and sandy clays 06m le B le 15m 1m le zemb le 25m

Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)

qult = function (qc and foundation shape)

Relationship shown in Figure A7 for square and strip footings

Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)

qult = Rkmiddot(qc - σvo) + σvo

Factor Rk obtained from Figure A8 accounts for surface and deep footings Separate Rk factor for intact and jointed clays

= (qc1middotqc2)05 qc

where qc1 is geometric mean from bearing elevation to 05B deeper and qc2 is geometric mean from 05B to 15B beneath foundation base

Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method

(Frank amp Magnan 1995)

Footing a surface

qult = kc middotqc + σvo

where kc = 032

Embedment case

Note Methods above generally consider undrained bearing capacity

B-9

Figure B7 Direct relationship between ultimate bearing stress in clays and measured cone tip

resistance (Schmertmann 1978)

B-10

Figure B8 Direct relationship between ultimate foundation bearing stress in clays and net

cone tip resistance (after Tand et al 1986)

B14 Direct CPT Assessments of Settlement and Displacements of Shallow Foundations

Methods for evaluating the magnitude of foundation displacements (s) can be found using

elastic theory subgrade reaction methods spring models and numerical simulations The

classical approach to footing settlement calculations on sands is to utilize elastic theory

solutions (eg Poulos amp Davis 1974) which take on the form

119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904

where q = applied footing stress and I = displacement influence factor from elasticity theory

The value of I depends upon foundation geometry footing rigidity embedment depth variation of soil

modulus with depth compressible layer thickness and other factors as discussed by Mayne amp Poulos

(1999) For instance for a flexible circular footing of diameter B resting on an infinitely deep soil

formation having a constant modulus Es with depth the influence factor I is equal to 1 for the center

point The results of in-situ field tests such as pressuremeter tests (PMT) standard penetration tests

(SPT) cone penetration tests (CPT) flat dilatometer tests (DMT) and other methods can be used to

ascertain the input value of soil modulus Es

Alternatively direct CPT methods have been investigated to provide a one-step assessment of

foundation displacements Selected methods for direct CPT evaluation of shallow foundation

settlements on granular soils is include DeBeers amp Martens (1957) Meyerhof (1965) DeBeer (1965)

Thomas (1968) Schmertmann (1970) Meyerhof (1974) Berardi amp Jamiolkowski amp Lancellotta (1991)

Robertson (1991) Lutenegger amp DeGroot (1995) Lehane amp Dougherty amp Schneider (2008) Gavin amp

Adekunte amp OrsquoKelly (2009) Mayne amp Illingworth (2010) and Uzielli amp Mayne (2012)

B141 Large Scale Footing Load Tests

The measured load-displacement response of shallow foundations is conducted in the field using either

stepped loading procedures or continuous rate of displacement methods Results from the well-

documented test program at the Texas AampM University (TAMU) site (Briaud amp Gibbens 1999 Briaud

2007) for five spread footings on sand are shown in Figure B9 Each footing clearly shows a nonlinear

B-11

behavior to loading over the range of testing This site is one of the national geotechnical

experimentation sites (NGES) funded by the National Science Foundation (NSF) FHWA and American

Society of Civil Engineering (ASCE)

Figure B9 Measured load-displacements for each of the five large footings at TAMU (Briaud

amp Gibbens 1994) sand site

B142 Generalized Direct CPT Method for Footing Response on Soils

The use of applied foundation stress versus normalized displacement curves (q vs sB) is generalized to

footings on sands silts intact clays and fissured clays A square root plotting of the normalized

displacements (sB) permits a single parameter characterization for each specific soil type that in turn

correlates with the net cone tip resistance The generalization is based on a statistical review of data

from large foundations (B gt 05m) involving 70 full-scale load tests with available CPT data For fine-

grained soils the data are from electronic piezocone tests where the proper correction from qc to qt has

been obtained to allow the best possible results The methodology permits an evaluation of the load-

displacement-capacity response of shallow square footings based on CPTs performed in the four types

B-12

of ground conditions For the TAMU footings the characteristic stress-normalized displacement

response is shown in Figure B10 where all five foundations can be represented by a single curve

Figure B10 Characteristic nonlinear stress-normalized displacement curve for the five TAMU

(Briaud amp Gibbens 1994) footings

B143 Characteristic Stress-Displacement Curves

Fellenius amp Altaee (1994) recommended the concept of a characteristic stress (q) vs normalized

displacement (sB) curve for foundation response on a given soil formation later supported by Decourt

(1999) Briaud amp Gibbens (1999) Lutenegger amp Adams (2003) and Briaud (2007) In this approach the

measured load (Q) vs displacement from individual footings of various sizes that rest on the same soil

conditions collapse to a unified relationship given in Equation B2

119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13

where af and bf are empirical fitting coefficients (Decourt 1999 Uzielli amp Mayne 2011 2012)

In a review of measured load tests data from large spread footings situated on different sands it has

been suggested that Equation B2 can be reduced to a single parameter expression by use of square root

plotting where the exponent bf is equal to 05 (Mayne amp Illingworth 2010 Mayne et al 2012)

119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3

where rs = fitted soil parameter from best fit line regression In the case of sands the coefficient rs

represents the supporting soil conditions including particle size relative density and fines content as

well as geologic origin aging and overconsolidation effects

The results from the five TAMU footings are presented in this format in Figure B11 that shows the

formation factor rs = 486 MPa for this sand deposit This single coefficient can be used to express the

nonlinear load versus settlement of any size footing on this sand site Moreover one criterion from

Eurocode for bearing capacity that is used by the European community identifies that stress (or load)

which corresponds to (sB) = 10 That is when the displacement equals ten percent of the footing

width then the capacity has been reached Therefore adopting this criterion for footings on sands

the ultimate bearing stress (qult) for footings on sand can be taken when (sB) equals 010 or when

square root of (sB) equals 0316 In that case this gives qult equal to 0316 middot rs For the TAMU site this

gives qult equal to 0316 times (486) which equals 153 MPa as illustrated by Figure B12

B-14

Figure B11 Characteristic stress versus square root normalized displacements for TAMU

footings

B-15

Figure B12 Eurocode or LCPC criterion applied to determine bearing capacity of TAMU sand

B15 Footing Database

A database of spread footings and large plates was assembled which considers only full-size shallow

foundations (05 le B le 6 m) that rest on sands silts and clays Sites for the foundations were also subjected to CPT The inclusion solely of large footings are significant because results from small-scale

model footings exhibit scale effects In fact experimental centrifuge work and numerical simulation

studies have shown that bearing capacity factors are size-dependent especially for factor Nγ decreasing

with footing width B (Kimura et al 1985 Cerato amp Lutenegger 2007 Mase amp Hashiguchi 2009) A

direct approach based on full-size footings provides a more reliable means for foundation evaluation

The compiled database is given in Table B3 with a total of 70 large footings and plates These include a

listing of 34 foundations on 13 sands 11 footings on 4 silt deposits 13 footings or large plate load tests

on 6 intact clays and 12 foundations on 5 fissured clays Most of the footings were square or nearly

square (80) while the remaining were circular The largest foundation was a mat 5 m by 14 m in plan

loaded by stacking Kentledge concrete blocks to cause a bearing failure in the underlying soft clay For

sands the largest footing consisted of the 6-m square concrete pad In terms of equivalent square

footings mean footing sizes were 155 m (sands) 114 m (silts) and 126 m (clays)

B-16

Additional details on the individual load tests and site conditions are given elsewhere (Mayne 2009

Mayne amp Illingworth 2010 Uzielli amp Mayne 2011 2012 Mayne et al 2012 Mayne amp Woeller 2014)

Table B3 Summary of large footings soil conditions and reference sources of database

Sand Site Location Soil Conditions Footings Numbers

Shapes and Sizes

ReferencesSource

College Station

Texas Pleistocene sand 5 Square 1 15 25 33 m Briaud amp Gibbens (1999)

Kolbyttemon Sweden Glaciofluvial sand 4 Rect B = 06 12 17 24 m

Bergdahl et al (1985 1986)

Fittija Sweden Glaciofluvial sand 3 Rect B = 06 17 24 m Bergdahl et al (1984 1985)

Alvin West Texas Alluvial sand 2 Circular D = 235 m Tand et al (1994)

Alvin East Texas Alluvial sand 2 Circular D = 22 m Tand et al (1994)

Perth Australia Silceous dune sand 4 Square B = 05 and 10 m

Lehane (2008)

Grabo T1C Sweden Compacted sand fill

1 Square B = 046 m Long (1993)





Labenne France Dune sand 6 Square B = 07 and 10 m

Amar et al (1998)

Green Cove Florida Brown silty sand 1 Circular D = 182 m Anderson et al (2006)

Durbin South Africa

White fine sand 1 Square B = 609 m Kantley (1965)

Porto Portugal Residual clayey sands

1 Circular D = 12 m and 1 plate

Viana da Fonseca (2003)

Jossigny France Soft clayey silt 2 Square B = 1 m Amar et al (1998)

Tornhill Sweden Glacial Baltic till 3 Square B = 05 1 2 m Larsson (2001)

Vagverkel Sweden Stiff medium silt 3 Square B = 05 1 2 m Larsson (1997)

Vattahammar Sweden Brn-Gry layered silt 3 Square B = 05 1 2 m Larsson (1997)

B-17

Table B3 Continued

Clay Site Location Soil Conditions Footings Numbers Shapes and Sizes

ReferencesSource

Baytown Texas Fissured Beaumont 2 plates with d = 076 m Stuedlein amp Holtz (2008)

Belfast Ireland Soft clay silt ldquosleechrdquo

1 Square Pad B = 20 m Lehane (Geot Engr 2003)

Bothkennar Scotland Soft silty clay 2 Square Pads B = 22 24 m

Jardine et al (1995)

Bangkok Thailand Soft to stiff clay 4 Square 067 075 090 10 m

Brand et al (1972)

Haga Norway Stiff OC clay 2 Square Footings B = 10 m

Andersen amp Stenhammar (1982)

Rio Grande Brazil Sandy residual clay 3 Square 04 07 and 10 m

Consoli et al (1998)

Shellhaven England Soft estuarine clay Rectangular 5 m by 14 m Schnaid et al (1993)

Texas City A Texas Coast Fissured Beaumont 3 circular plates d = 058 m

Tand et al (1986)

Texas City B1 Texas Coast Fissured Beaumont 2 circular plates d = 058 m

Tand et al (1986)

Texas City B2 Texas Coast Fissured Beaumont 1 circular plate d = 058 m

Tand et al (1986)

Alvin Texas Texas Coast Fissured Beaumont 3 circular plates d = 058 m

Tand et al (1986)

For the series of footings on sands Figure B13 shows the summary of measure formation factors (rs) for

the 34 foundation load tests Also indicated on the graph are the corresponding mean qc values of the

sands averaged over 15middotB beneath the bearing elevations

Note also in sands that the qt and net resistance (qnet) are all very close to the measured qc since (a)

porewater pressure corrections for the a-net value are negligible in clean sands and (b) overburden

stresses are typically only 1 to 3 of the magnitude of qc This is especially true of shallow depths

Therefore for clean sands qnet is approximately qt which is approximately qc

B-18

Figure B13 Summary of sand formation factors with corresponding CPT resistances

As suggested by Briaud (2007) various in-situ measurements can be used to normalize the results from

cone penetration tests in this case Herein the qc from the CPT soundings in the sands are used to

normalize the footing stress axis as presented in Figure B14 It is evident that the results of the load-

displacement response of all 34 footings on 13 different sands can be captured by the characteristic

stress versus normalized displacement with the inclusion of the site-specific cone tip resistance In this

case the mean trend for all footings and sand sites indicates a simple expression shown in Equation B4

119954119943119952119952119957119946119951119944 119956 Clean quartz-silica sands = 120782 120787120790120787radic B4

119954119940 119913

B-19

Figure B14 Normalized foundation stresses vs square root of normalized displacement for

spread footings on sand (modified after Mayne et al 2012)

The results of measured force-displacement curves (Q vs s) from footing load tests are nonlinear and

thus the definition of capacity must be addressed Kulhawy (2004) identified three main curve types

that occur during load testing depicted in Figure B15 The most common response (Type A) shows no

clear peak and is observed in sands and silts as well as slow loading of insensitive clays and fissured

clays In this case ldquocapacityrdquo can be defined by the Eurocode corresponding to the load (or stress) when

sB=10 (Amar et al 1998) For foundations situated on many clays a plateau capacity is reached as

depicted as Type B response In rare cases where sensitive clays or structured soils are encountered a

Type C relationship occurs whereby the load-displacement curve reaches a peak followed by strain

softening In the one case observed in this study for Type C loading (Haga clay) the corresponding

ldquopeak capacityrdquo was reached at a pseudo-strain of sB = 4 as shown in Figure B16 followed by some

strain softening In the cases of soft clays at Belfast and Bothkennar a plunging type (plateau failure)

was achieved at around sB=7

B-20

Figure B15 Definitions of capacity from three types of observed foundation load test

responses (modified after Kulhawy 2004)

B-21


strains (sB) for all footings

In the case of fine-grained soils that develop excess porewater pressure during cone penetration the

use of the net total cone resistance (qtnet) which equals the cone tip resistance minus the total stress

must be considered (Lunne et al 1997) As noted earlier the correction of CPT data in sands is minimal

because of the low value of penetration porewater pressures and the fact that total overburden stress is

small relative to the cone resistance especially for shallow soundings Thus the use of qtnet may be

substituted for qc into the trend of Figure B1

B16 Undrained and Drained Capacity

The footings on sands typically show drained response with no excess porewater pressures developed

during loading For the foundations situated on silts analyses by Larsson (1997 2001) concluded that

these too show drained conditions For shallow foundations on clays the entire range of drainage cases

are possible ranging from undrained to partially-drained to fully-drained depending upon the applied

rate of loading relative to the permeability of the geomaterial If sufficient data were available for each

of the clay sites then the recent evaluation of drainage condition could be assessed using normalized

velocity criteria (Chung et al 2006) However this was not possible with the current data set Instead

B-22

the test loadings of footings on clays have essentially been assumed to primarily occur under undrained

conditions following recommendations by Tand et al (1986) For this case a limiting bearing stress can

be calculated from bearing capacity analyses and related directly to the CPT readings

From classical bearing capacity calculations using limit plasticity solutions the ultimate foundation stress

is shown in Equation B5

= 119925119940 ∙ 119956119958 B5 119954119958119949119957

where Nc equals the bearing factor for constant volume (514 for strip and 614 for square or circular

plan) and su equals the undrained shear strength of the clay For the case of soft to firm clays of low to

medium sensitivity the operational strength can be related to the preconsolidation stress (Mesri 1975

Jamiolkowski et al 1985) as shown in Equation B6

prime 119956119958 = 120782 120784120784120648119953 B6

Finally the effective preconsolidation stress has been linked directly to the net cone resistance (Chen amp

Mayne 1996 Demers amp Leroueil 2002) as shown in Equation B7

prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7

Combining Equations (B5) (B6) and (B7) results in direct expressions for undrained bearing capacity on

intact clays from CPT results as shown in Equations B8 and B8-2

Strip footing 119954119958119949119957 = 120782 120785120789120785(119954119957 minus 120648119959119952) B8

B-23

Squarecircle 119902119906119897119905 = 0445(119902119905 minus 120590119907119900) B8-2

Equation (B8-2) is in excellent agreement with Tand et al (1986) database study for the case of intact

clays and footings with no embedment Their study also provided a lower relationship for foundations

situated on jointed clays specifically recommending that the bearing stress not exceed 030 qtnet

Review of the available data in Figure 3 shows this to be rather conservative and a more realistic value

may be taken at 040 qtnet for fissured clays

B161 Normalized Undrained Footing Response

The characteristic stress vs normalized displacement curves for footings on clays exhibiting undrained

conditions can be verified using a relatively recent case study on Beaumont clay by Stuedlein amp Holtz

(2010) The Baytown Texas site was investigated using an exploration program of soil borings

piezocone soundings and laboratory testing The field testing included a large square footing (B = 276

m) and two small circular plates (D = 076 m) The measured load-displacement responses for all three

foundations are presented in Figure B17 Of course the large footing carried considerably higher loads

than the two small plates on the same soil deposit Yet using the aforementioned characteristic stress

(q) vs square root of normalized displacement (sB) essentially gave a unique relationship for all 3

foundations as presented in Figure B18 In this case utilizing the form of Equation B2 the

characteristic coefficient rs is 177 MPa for this deltaic clay formation

B-24

Figure B17 Measured load vs displacement curves for 3 shallow foundations on clay at

Baytown Texas (Stuedlein amp Holtz 2010)

B-25

Figure B18 Characteristic stress vs square root of normalized displacement for all three

foundations at Baytown site (Stuedlein amp Holtz 2010)

B17 General Direct CPT Method

In a manner analogous to the normalization of applied foundation stresses shown in Figure B14 for

footings on sands a generalized direct CPT method for shallow foundations on different soils can be

made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9

where qcapacity is equal to the foundation bearing capacity of the ground and hs is equal to the empirical

fitting term that depends on soil type Specifically hs equals 058 for sands 112 for silts 147 for

fissured fine-grained soils and 270 for clays A summary of the normalized stress versus the normalized

displacement curves for these four soil types is presented in Figure B19 where the data fitting to obtain

the hs parameter on clays are limited to (sB) lt 004 corresponding to undrained loading The data on

B-26

silts and sands are considered fully drained whereas the fissured clay subset may be partially drained to

undrained The statistical measures for obtaining the fitted parameter hs are quite good as shown in

Figure B20 with the coefficient of determinations (r2) values of 0947 for sands 088 for silts 0935 for

fissured and 0925 for intact clays Additional statistical evaluations on the database are provided by

Uzielli amp Mayne (2011)

Figure B19 Normalized foundation stress vs square root of normalized displacements

B-27

Figure B20 Applied foundation stress vs CPT-calculated stress for 67 footings

The same data are shown on Figure B21 in a log-log plot to emphasize the early parts of the load-

displacement responses of these footings The best fit line from regression analyses shows excellent

statistical correlations As detailed earlier the undrained bearing capacity of square or circular footings

on clays can be taken as approximately 040 times qtnet For silts and sands that experience drained

loading with no excess porewater pressures being developed and no clear peak value for capacity the

(sB) =10 criterion can be used In this case Equation 7 can be used directly to obtain stress q equal to

qcapacity when (sB)05 is 0316

An alternate approach is presented in Figure B22 summarizing the direct CPT method for estimating the

load-displacement-capacity of shallow square footings on four soil types The maximum values of soil

bearing capacity (qmax) can be capped at stresses corresponding to a percentage of the average

measured qtnet determined over the depth interval from the foundation bearing elevation to 15middotB

below the foundation In such an approach the foundation bearing capacity can be taken as

(qcapacityqtnet) of 02 for sands 035 for silts 040 for fissured and 045 for intact clays Using the assigned

values of the hs parameter the corresponding cut-off for the general stress-normalized displacement

relationship given by Equation 7 can be established specifically giving corresponding values of the

B-28

normalized displacements which can also be considered as pseudo-strains equal to (sB) max of 4 for

clays 7 for fissured clays 10 for silts and 12 for sands

Figure B21 Applied footing stress vs CPT calculated stress on logarithmic scales

B171 CPT Soil Material Index Ic

The soil behavioral type can be assessed indirectly by CPT using a material index (Ic) as defined by

Robertson (2009a b) shown in Equation B10

119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29

Figure B22 Summary graph and equations of direct CPT method for evaluating the vertical

where Qtn equals the stress-normalized cone tip resistance and Fr equals the normalized sleeve friction

calculated from the cone penetrometer readings as shown in Equations B11 and B12 respectably

(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)

where σatm equals a reference stress equal to atmospheric pressure (σatm = 1 atm asymp 1 bar asymp 100 kPa) In

the initial evaluation the exponent n is set to n = 1 to find the soil behavioral type (SBT) based on a 9-

zonal chart as shown in Figure 23 The value of exponent n varies with soil type ranging from n = 1 in

intact clays and decreasing with increasing grain size to around approximately 075 in silts and

B-30

approximately 05 plusmn 02 in clean sands The appropriate value of exponent n is found by iteration until

conversion using the relationship (Robertson 2009b) shown in Equation B13

prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950

As the distinction that footings on intact clays subjected to fast loading are behaving primarily under

undrained conditions (ie constant volume) Robertson (2009a) suggested that this occurs when Ic gt 27

while in contrast Ic lt 25 corresponds more or less to drained loading (ie no excess porewater

pressures)

The CPT material index can be used to identify soil type and the corresponding characteristic hs value for

estimating footing load-displacement response In a number of the case studies investigated the results

of the sleeve friction readings were also available to allow the calculation of Ic with depth at these sites

Figure B23 shows the tentative relationship between the values of the hs parameter plotted versus the

corresponding CPT material index with an approximate trend given by Equation B14

120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31

Figure B23 Foundation soil formation parameter hs versus CPT material index Ic

Soil behavioral type from the Ic ranges given by Robertson (2009b) and are shown in Figure B23 with an

overall general agreement with the known soil classifications In summary the CPT can provide an

evaluation of the load-displacement-capacity response directly via Equation B15 which may be of

interest in mixed soil types such as silty sands sandy silts and the like

120784120785 Footing stress 119954 = 119954119951119942119957 ∙ radic(119956frasl119913) ∙ [120784 120790 minus

)120783120787] B15

120783+(119920119940frasl120784120786

As discussed earlier foundation bearing capacity may be taken by a limiting value of pseudo-strain (sB)

max or by qmax as specified as a percentage of qtnet This definition of bearing capacity can be seen in

comparison to soil type as seen in Figure B24

B-32

Figure B24 Bearing capacity ratio qmaxqnet versus CPT material index Ic

B172 Rectangular Foundations

This same elastic solution from Giroud (1968) for a CPT method for square and circular foundations were

utilized for rectangular foundations The study covered rectangular distortions (AB) ranging from 1

(square) to very long foundations with AB equal to 20 where A represents the foundation length and B

represents the foundation width (Mayne amp Dasenbrock 2017) The influence factor for rectangular

shaped foundations is given in Figure B25 and the expression in Equation B16

= (119912119913)120782120785120786120787 119920119912119913 B16

B-33

Figure B25 Influence factor for rectangular foundations from elastic theory solution (Mayne

amp Dasenbrock 2017)

Including 32 full scale load tests on sands results from an additional 98 shallow foundations have been

reported in previous studies The foundations were primarily square or circular and among these include

large spread footings with measured settlements and bearing capacity Table B4 shows a summary of

the number of footings and footing sizes used in the study The range of length to width ratios varied

from 1 lt (AB) lt 23 with an average value of (AB) equal to 238 The embedment depth to footing

width ranged from 0 lt DeB lt 222 and averaged 046

Table B4 Sources for shallow foundation performace

Source of Data No of

Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027

The solution for shallow foundation settlements given back in Equation B1 combined with

the expression in Equation B16 takes on the form

119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956

Combining the direct CPT equation developed from the 32 full scale load tests (Equation B15)

together with the influence factor for rectangular shaped foundations becomes

minus120782120785120786120787 120784120785 119912 Footing stress 119954 = 119954119951119942119957 ∙ radic(119956frasl119913) ∙ [120784 120790 minus

)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions

A direct CPT method for square rectangular and circular shallow footings is developed using a database

of 166 full-scale field load tests where the minimum footing size of an equivalent square width of 05 m

is established to avoid scaling issues The use of load vs displacement is generalized by characteristic

stress vs normalized displacement (sB) and further simplified by a square root plotting technique that

captures the ground response in a single parameter that is related to the qtnet Data are grouped

according to four main soil categories sands silts fissured clays and intact clays It is generally believed

that the footings on sands and silts exhibit fully drained behavior while in intact clays an undrained

response occurs under conditions of constant volume

B-35

The summary equation for evaluating the vertical stress-displacement-capacity of square rectangular

and circular footings is given by

minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913

where the parameter hs also relates directly to Ic The foundation capacity depends upon the mode of

failure as well as drainage conditions and can be taken as a fraction of qtnet where qmaxqtnet is 020 in

sands 035 in silts 040 in fissured clays and 045 in intact clays as shown in Figure 24 Alternatively the

capacity can be defined using a limiting pseudo-strain given by (sB) max of 4 in clays 7 in fissured

10 in silts and 12 in sands

B-36

APPENDIX C

List of Figures

Figure C1 Components of Axial Pile Capacity C-3

Figure C2 Selected alpha relationships as function of the undrained shear strengthC-5

Figure C3 Pile side friction coefficient β in terms of effective friction angle and overconsolidation ratio

C-11

Figure C4 Concept of Direct versus Rational Method for CPT evaluation of axial pile capacityC-14

Figure C5 Soil behavior type using CPT via original UniConeC-19

Figure C6 Soil behavior type using CPT via Modified UniCone C-19

Figure C7 Nine-zone soil behavioral soil type using normalized piezocone parameters C-21

Figure C8 Summary of Modified UniCone Method for direct CPT assessment of axial pile capacity C-22

Figure C9 Elastic continuum solution for axial pile response under compression and tension loadingC-24

List of Tables

Table C1 Alpha Methods for Axial Pile Side Friction where fp = αmiddotsu C-5

Table C2 Beta Methods for Axial Pile Side Friction where fp = βmiddotσvoC-9

Table C3 Selection of Direct CPT Methods for Axial Pile CapacityC-15

C-1

C1 EVALUATING DEEP FOUNDATION RESPONSE FROM CONE

PENETRATION TESTS

C11 Introduction

The axial response of deep foundations includes the load-displacement-capacity and axial load transfer

when driven pilings and drilled shafts are loaded in compression and uplift The use of cone penetration

testing (CPT) especially seismic piezocone tests (SCPTu) are advantageous since they provide

information on the subsurface soils and their geomaterial properties The SCPTu provides at least four

measurements on soil behavior with depth including (a) cone tip resistance qt (b) sleeve friction fs (c)

penetration porewater pressure u2 and (d) shear wave velocity Vs This offers the opportunity to

determine the geostratigraphy unit weight effective overburden stress shear strength stress state

and stiffness of the ground from a single exploratory sounding thus values in economic expediency

and reliability The axial pile capacity can be calculated based on static equilibrium of forces acting along

the sides and base of the pile foundation The displacement of the pile can be ascertained using

elasticity theory from closed-form solutions boundary elements andor finite element analyses Load

transfer occurs along the length of the pile and only a portion of axial forces are transmitted to the base

or toe or tip of the pile These too can be assessed using elasticity solutions

An alternate approach to pile capacity involves the utilization of direct CPT methods available

since the 1970s but in the past 10+ years a number of new and statistically reliable algorithms

have been developed which can be implemented for highway design and construction

C111 Axial Pile Capacity

The axial compression capacity of a single pile foundation is composed of a shaft or side component and

end-bearing component at the base as depicted in Figure 1 For a circular pile the side capacity (Qs) is

determined from the unit side friction (fp) acting along the surface area of the shaft which is As = πmiddotdmiddotL

where d = pile diameter and L = length embedded below grade

If the magnitude of side friction is uniform and constant with depth the side capacity is simply

C-2

119928119956 = 119943119953 middot 119912119956 C1

Moreover however many piles extend through multiple layers and a summation of unit side

frictions acting on various pile segments must be tabulated over the length of the pile as

suggested by Figure C1

Figure C1 Components of Axial Pile Capacity

The unit-end bearing resistance (qb) acts over the base of the pile tip where the area of a circular pile is

given by Ab = πmiddotd24 For piles in compression loading the base capacity is determined from Equation

C2

119876119887 = 119902119887 middot 119860119887 C2

and for piles in tension (or uplift) it is normally taken that Qb = 0

C-3

C112 Pile Unit Side Friction

Several different approaches can be adopted for evaluating the unit pile side friction (fp) prior to full-

scale load testing and construction (Poulos amp Davis 1980 ONeill 2001) The most common types

including the alpha and beta methods as summarized in Tables 1 and 2 respectively In the alpha

method an empirical coefficient (α) is applied to the undrained shear strength (su) of clay soils to

obtain

119891119901 = 120572 middot 119904119906 C3

An illustrative example of alpha curves is shown in Figure C2 A difficulty with the alpha

method is the evolution of many variants and changes to the expressions for its estimation It is

also restricted in its use for specific types of piles in clays and fine-grained soils

C-4

Figure C2 Selected alpha relationships as function of the undrained shear strength

Table C1 Alpha Methods for Axial Pile Side Friction where fp = αmiddotsu

C-5

Reference

Source

Alpha Equation Remarks

Tomlinson

(1957)

2716801102

uu ssAll pile types (steel

concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)

Kerisel α = 07 - 031middotln(su)

Peck α = 10 - 02middotsu

Woodward α = 091middot(su)091

Driven piles in clay

Note su in ksf

Semple

(1980) )1(511

)1(1

OCR

OCR Summary from 9 series of

pile load tests

American

Petroleum

Institute (API

1981)

For suσvo le 035 α = 10

For suσvo gt 080 α = 05

Otherwise α = 1 + 1111middot (035 - suσvo)

Driven steel pipe piles

Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100

where 75 lt su (kPa) lt 210

1 Driven piles

2 Bored piles

3 Driven piles in till with L

lt 10middotd

API (1987) 1 α = 10 for su lt 25 kPa Driven piles in clay other

than Gulf of Mexico

C-6

2 α = 125 - 001middotsu for 25 lt su lt 75 kPa

3 α = 050 for su gt 75 kPa

Kulhawy amp

Jackson

(1989)

α = 021+026middot(σatm su) le 1 Analyses of 106 drilled

shaft foundations

Table C1 Continued

API (1989) 05 For suσ vo le 1 α =

su radic frasl prime σvo

minus025 su For suσ vo gt 1 α = 05 ∙ ( frasl prime ) σvo


Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin

Λ = (1-CsCc) asymp 080 for

insensitive clays

Nowacki et al

(1996)

minus05 su For suσvo le 07 α = 05 ∙ ( frasl prime ) σvo

minus02 su For suσvo gt 07 α = 055 ∙ ( frasl prime ) σvo

Driven pile foundations

Kolk and van

der Velde

(1996)

α =09middot FL middot (suσvo)-03 lt 10

FL = [ (L - z)d]-02 = length term

L = pile length

Note su obtained from

UU lab tests

Applicable to driven piles

in clays

C-7

Knappett amp α = 1 for su le 30 kPa Non-displacement piles in

Craig (2012) fine-grained soils α = 116 - (su185) for 30 kPa le su le 150 kPa

α = 035 for su gt 150 kPa

Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)

∙ su ( frasl prime σvo

minus03

) Displacement piles in fine-

grained soils

z = depth at point considered

d = outside diameter of pile

Karlsrud et al

(2005 NGI

Method)

α = function (suσvo and plasticity index) Driven pilings

Fleming et al

(2009) 05 minus05 su su For su σvo le 1 120572 = ( frasl prime ) ∙ ( frasl prime )

σvo NC σvo

05 minus025 su su For su σvo gt 1 α = ( frasl prime ) ∙ ( frasl prime ) σvo NC σvo

For driven piles in clay

where (suσvo)NC is the

normalized shear strength

for normally-consolidated

clay

Brown et al

(2010)

α = 0 for 0 lt z le 5 feet

α = 055 for z gt 5 feet and (suσatm) le 15

α = 055-01middot(suσatm -15) for 15le (suσatm) le 25

Drilled Shafts and Bored

Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)

α = function (su σvo and plasticity index) Driven pilings

Note as reported by Karlsrud (2012)

In the beta method the coefficient β is applied to the effective overburden stress (σvo) at the point of

concern along the pile length

prime = 120573 C4 119891119901 middot 120590119907119900

Table C2 Beta Methods for Axial Pile Side Friction where fp = βmiddotσvo

Reference Source Beta Equation Remarks

Burland (1973) β = (1-sinϕ) middot tan ϕ Piles in NC clays

Meyerhof (1976) β = Kmiddottanδ where K = K0 = for NC clays

K = 15middotK0 for OC clays

Poulos amp Davis

(1980)

β = (1-sinϕ) middot tan ϕ middot OCR05 Piles in OC stiff clays

Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)

β = (1-sinϕ) middot tan ϕ middot OCRsinϕ Piles in quartz sands and insensitive

clays

ONeill (2001) β = (1-sinϕ) middot tan θ middot OCRsinϕ Drilled shafts where θ = (δϕ)middotϕ and

δ = interface friction between pile and soil

Fleming et al

(2009)

β = Kmiddottanδ K = lateral stress coefficient and δ =

friction angle between soil and pile

material

Karlsrud (2012) β = fctn (OCR and PI) PI = plasticity index of the clay ()

Mayne and Niazi

(2017)

β = CM middot CK middot K0 middot tanϕ

where K0 = (1-sinϕ) middot OCRsinϕ for soils

that are virgin loaded then unloaded

CM = pile material factor = 1 (drilled

augered) 09 (prestressed or precast

concrete) 08 (timber) and 07

(steel) and CK = pile installation factor

= 09 (bored or augered) 10 (low

displacement eg H-pile or open-end

pipe) and 11 (driven high

displacement eg prestressed

concrete closed-end pipe)

As the beta approach applies to all types of soils (gravels sands silts and clays) and more or

less has remained unchanged since its advent circa 1970 it has been selected for further

discussion herein In the direct form for calculation of unit pile side friction the expression is

given by

119874119862119877119904119894119899120601 prime prime 119891119901 = 119862119872 ∙ 119862119870 ∙ (1 minus 119904119894119899120601prime) middot ∙ 119905119886119899120601prime ∙ 120590119907119900 C5

C-10

where CM is a soil-pile interface friction coefficient and CK = pile installation factor Values of CM are in

the range 07 lt CM lt 10 depending upon pile material (steel wood concrete) and ranges for CK vary

09 lt CK lt 11 depending upon method of installation (auger drill driven) as detailed in Table 2

The value of K0 is limited to the passive stress coefficient which for the simple Rankine case is given by

KP = (1+sin ϕ) (1 - sin ϕ ) Thus there is a maximum value of overconsolidation ratio for which

Equation C5 applies given by OCRlimit = [(1+sin ϕ ) (1 - sin ϕ )2] (1sinϕ)

The corresponding graph in Figure C3 shows the relationship for β in terms of ϕ and OCR


C113 Pile Unit End Bearing

C1131 Theoretical Considerations

The evaluation of the end bearing resistance of pile foundations is commonly determined using

limit plasticity theory where

prime Drained loading = C6 119902119887 119873119902 middot 120590119907119900

C-11

Undrained loading 119902119887 = 119873119888 middot 119904119906 C7

where the value of σvo is calculated at the depth z = L and the value of su is the average undrained shear

strength from z = L to a depth z = L + d beneath the pile tip For a circular pile the limit plasticity solution

for undrained loading gives (Vesic 1977) a value Nc = 933 while a deep strip foundation would employ a

value of Nc = 824 For drained loading the expression for Nq for a deep foundation is given by (Vesic

1977)

)]arctan()sin1(tan21[)](tan1[sin1

sin1)tanexp( 2 BLABNq

C8

where A and B are the pile plan dimensions (for a circular pile A = B) and L = pile length For a

circular pile this can be approximated by

asymp 077(120601prime75deg) 119873119902 C9

over a range of effective friction angles 20deg le ϕ le 45deg

C1132 Practical Considerations

For drained loading of sands the full calculated end bearing capacity will never be realized because it

would require the pile to move a distance equal to its diameter This is beyond practical use and

therefore the limit plasticity solutions must be clipped to a fraction of the calculated value Another

reason for using a reduced end-bearing capacity is due to strain incompatibility since the side capacity is

mobilized early while the end-bearing is engaged much later So to have compatible values of Qs and

Qb the qb must be reduced using the following guidelines (Randolph 2003 Mayne 2007)

prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909

where fx = strain incompatibility factor

fx = 010 for drilled shafts augered cast-in-place and bored piles

fx = 020 for driven low displacement piles (opened-ended steel pipe and H-piles)

fx = 030 for driven high-displacement piles (ie solid piles PSC and closed-ended pipe)

C-12

For undrained loading of circular piles in compression no reduction of end-bearing is necessary

therefore

119902119887 = 933 middot 119904119906 C11

C12 Direct CPT Methods for Pile Capacity

In the direct CPT method the penetrometer readings are scaled directly via specified algorithms to

obtain the pile unit side friction and end-bearing as depicted in Figure C4 As many as 40 different

direct CPT methods have been developed over the past five decades as summarized by Niazi amp Mayne

(2013) Starting circa 1970 many of these early methods relied on hand-recorded data from field

mechanical CPTs where only qc data were obtained at 20 cm intervals or later with mechanical readings

of both qc and fs using special sets of inner and outer rods that recorded vertical load for tip and loads

for tip plus sleeve in alternating increments Also early pile load test data were often obtained from

top-down measurements of load-displacement

C-13

Undrained qb = Nc su

Qside = S (fp DAs)

QTotal = Qs + Qb - Wp

fp = cmck Ko svorsquo tanrsquo

Qbase = qb Ab

Drained qb = Nq svorsquo

Method Two Rational or ldquoIndirectrdquo Method

Method OneldquoDirectrdquo CPT Method

(Scaled Pile)

fp = fctn (soil type pile

type qt fs and u2)

qb = fctn (soil type qt-u2)

qb = unit end bearing

unit sidefriction fp

OCR su Ko t DR rsquo

AXIAL PILE CAPACITY FROM CPT READINGS

Figure C4 Concept of Direct versus Rational Method for CPT evaluation of axial pile capacity

Beginning in the mid-1990s as the modern electric piezocone (CPTu) was implemented the newer

equipment offered improved data and better resolution because of the additional reading of porewater

pressures and correction of raw measured qc to total resistance qt In addition the use of electronic

digital data collection and field computers proved superior in field measurements and recordings As a

consequence several reliable CPT methods for axial pile capacity have been developed Moreover

parallel improvements in full-scale pile load testing occurred and now include modern strain gage

instrumentation digital data recording and automated testing procedures Also the testing can be

single direction (compression or tension) or bi-directional as in the Osterberg cell

C-14

Table C3 provides a selection of recent direct CPT methods that have become available over the

past two decades A number of these (namely ICP NGI UWA and Fugro) were funded by the

offshore industry because of the growth of oil amp gas reserves and windfarm installations thus

necessitating increased concerns on risk probability and reliability in the site investigations for

offshore platforms and design of large driven monopile foundations These direct CPT methods

are often statistically based on large datasets compiled from full-scale load tests made

worldwide (eg Schneider et al 2008)

Table C3 Selection of Direct CPT Methods for Axial Pile Capacity

Method Pile Types Soil Types References CPT

data

Additional

parameters

needed

Unicone Method all types Sands silts

clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)

fs and u2 No guidance given

on end bearing

resistance

Imperial College

Procedure (ICP)

OE and CE Sands Chow et al

(1997 PhD)

Jardine et al

(2005)

qt Interface friction

angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15

OE and CE Clays Jardine et al

(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)

OE and CE Sands Clausen et al

(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()

Fugro Method OE and CE Sands Kolk et al

(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt

OE and CE Sands Lehane et al

(2005)

qt IFR = infilling ratio

related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design

Enhanced Various Sands silts Niazi and qt u2 Based on 330 load

Unicone Method clays and Mayne (2015 and fs tests including

mixed soils 2016) bored augered

jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness

Australia) Clays Lehane et al

(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued

Note previously called Marine Technical Directorate (MTD)

C-17

Many of the offshore CPT methods relate to driven piles either in sand or clay as that is

common deep foundation for those purposes Of particular interest are the Unicone and

Modified Unicone Methods since they use all three readings of the piezocone (qt fs and u2)

and address a variety of pile foundation types

C13 Modified UniCone Method

The modified UniCone Method was developed based on a total 330 pile load tests which were

associated with SCPTu data during their site investigations (Niazi and Mayne 2015 2016) This

represents a threefold increase over the original UniCone database that was built upon data

from 106 pile load tests (Eslami amp Fellenius 1997)

For the original UniCone algorithms use is made of the effective cone resistance (qE)

119902119864 = 119902119905 minus 1199062 C12

and a chart of qE vs fs provided an approximate soil classification in five distinct groups as shown by

Figure C5 Later in the modified approach a better delineation of the larger dataset gave soil

subclassifications as indicated by Figure C6

C-18

Figure C5 Soil behavior type using CPT via original UniCone

Figure C6 Soil behavior type using CPT via Modified UniCone

C-19

In the modified approach the 9-zone normalized soil behavioral type (SBTn) is ascertained using CPT

data in conjunction with the Robertson (2009) charts as shown in Figure C7 As discussed in Appendix

A and B the SBTn method uses the normalized cone resistance (Qtn) normalized sleeve friction (Fr())

and CPT material index Ic This permits a much wider range in the calculated pile side friction because fp

is related as a continuous curve with Ic rather than only 5 values that are assigned in the original

scheme

The pile unit side friction (fp) is obtained from qE and the CPT material index Ic using the following

expression at each elevation along the sides of the pile

10(0732 middot 119868119888 minus 3605) fp middot middot C13 = 119902119864 middot 120579119875119879 middot 120579119879119862 120579119877119860119879119864

C-20

Figure C7 Nine-zone soil behavioral soil type using normalized piezocone parameters

(after Robertson 2009 Mayne 2014)

where θPT = coefficient for pile type (θPT = 084 for bored piles 102 for jacked piles 113 for driven

piles) θTC = coefficient for loading direction (θTC = 111 for compression and 085 for tension) and θRATE

= rate coefficient applied to soils in SBT zones 1 through 7 (θRATE = 109 for constant rate of penetration

tests and 097 for maintained load tests) Since CPT provides data at regular intervals of 2 cm to 5 cm

along the sides of the pile the average fp from z = 0 to z = L can be used directly in Equation 1 to obtain

the shaft capacity Qs

C-21

The pile end bearing resistance is obtained from

middot 10(0325middot 119868119888minus1218) 119902119887 = 119902119864 C14

where qE is averaged in the vicinity of the pile tip Figure C8 shows the Modified Unicone Method in

graphical format For sensitive clays of zone 1 please see additional discussions by Niazi amp Mayne

(2016)

Figure C8 Summary of Modified UniCone Method for direct CPT assessment of axial pile

capacity

C-22

C14 Axial Pile Displacement

The movement of pile foundations can be assessed using elastic continuum theory (Poulos amp

Davis 1980 Randolph 2003 Mayne amp Niazi 2017) These relationships have been developed

using finite element analyses boundary elements and analytical closed-form solutions For the

latter approach the top displacement of a rigid pile subjected to a vertical force is shown in

Figure C9 This gives the movement at the top of a pile subjected to either compression or

tension (uplift) loading Also the percentage of axial load transferred to the pile toe is

determined For piles extending through various soil layers the elastic solution can be

implemented by using a set of stacked pile segments each with its own stiffness as

represented by a soil Youngs modulus

For pile groups the use of computer software is recommended Several available programs can handle

pile groups under axial and lateral moment loading such as DEFPIG (Univ Sydney) and PIGLET (Univ

Western Australia) A full listing of pile foundation software is given at the Geotechnical amp

GeoEnvironmental Service Directory wwwggsdcom

Ps

= Sh

aft

Load

sL

t

tEd

IPw

Load Transfer to Base

)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb

Randolph Elastic Solution

Ground Surface

Top Load Pt = Ps + Pb

Rigid Pile Response

wt = displacementd = diameter

L = length

Es = Elastic Soil Modulus

Es0 (at z = 0) surface

EsM (at z = L2)mid-length

EsL (at z = L)full length

Load Transfer to Shaft

E = EsMEsL = Gibson parameterE = 1 for homogeneous case (constant E)E = 05 for pure Gibson case (Es0 = 0)

where FCT = load direction (= 1 compression 0 tension)

21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23

Figure C9 Elastic continuum solution for axial pile response under compression and tension

loading

C-24

C15 Acknowledgements

The author appreciates the help of Fernando Illingworth of PonsEcuador Meena Viswanth of

GeoSyntecAtlanta and Dr Marco Uzielli of GeoRiskItaly in helping to collect and process the data

Funding support was provided by David Woeller of ConeTec Richmond BC

C-25

Structure Bookmarks

CONE PENETRATION TEST DESIGN GUIDE FOR STATE GEOTECHNICAL ENGINEERS

TABLE OF CONTENTS

LIST OF TABLES


CHAPTER 2 DIRECT CPT METHOD FOR SOIL CHARACTERIZATION

CHAPTER 3 DIRECT CPT METHOD FOR SHALLOW FOUNDATIONS


REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C

To request this document in an alternative format such as braille or large print call 651-366-4718 or 1-800-657-3774 (Greater Minnesota) or email your request to ADArequestdotstatemnus Please request at least one week in advance


1 Report No

MNRC 2018-32




Engineers

5 Report Date

November 2018 6

7 Author(s)







(C) 99008 (WO) 249
















Unclassified


Unclassified

21 No of Pages

225

22 Price



Prepared by

Ryan Dagger

David Saftner



Paul W Mayne



November 2018

Published by











TABLE OF CONTENTS



21 Introduction 4






241 Step 1 8

242 Step 2 9












31 Procedure 56












41 Introduction70


421 Step 172

422 Step 272

423 Step 373

424 Step 473




REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C


1 Report No

MNRC 2018-32




Engineers

5 Report Date

November 2018 6

7 Author(s)







(C) 99008 (WO) 249
















Unclassified


Unclassified

21 No of Pages

225

22 Price



Prepared by

Ryan Dagger

David Saftner



Paul W Mayne



November 2018

Published by











TABLE OF CONTENTS



21 Introduction 4






241 Step 1 8

242 Step 2 9












31 Procedure 56












41 Introduction70


421 Step 172

422 Step 272

423 Step 373

424 Step 473




REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C



Prepared by

Ryan Dagger

David Saftner



Paul W Mayne



November 2018

Published by











TABLE OF CONTENTS



21 Introduction 4






241 Step 1 8

242 Step 2 9












31 Procedure 56












41 Introduction70


421 Step 172

422 Step 272

423 Step 373

424 Step 473




REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C

TABLE OF CONTENTS



21 Introduction 4






241 Step 1 8

242 Step 2 9












31 Procedure 56












41 Introduction70


421 Step 172

422 Step 272

423 Step 373

424 Step 473




REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C










41 Introduction70


421 Step 172

422 Step 272

423 Step 373

424 Step 473




REFERENCES 87

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

LIST OF FIGURES








































LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C
















LIST OF TABLES





























1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C






















1













settlements

2


3

Symbol Parameter












CHARACTERIZATION

21 INTRODUCTION






4

Kʹ Bulk modulus








where ga = 98 ms2






5

22 SOIL UNIT WEIGHT



soils



1 120574119905 = 120574119908 ∙ [122 + 015 ∙ ln (100 ∙ 119891119904 + 001)]

120590119886119905119898









6







2

7

119891119904 119865119903() = 100 ∙

(119902119905minus120590119907119900)





3 (119954119957minus120648119959119952)120648119938119957119950 = 119951 119928119957119951 prime (120648119959119952120648119938119957119950)

4

prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 120590119886119905119898

119899 le 10

5 119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784




241 Step 1




8


242 Step 2







119876119905119899 lt 12119890119909119901(minus14 ∙ 119865119903)


1








9

Soil Type m

Fissured clays 11

Intact clays 10

Sensitive clays 09

Silt mixtures 085

Silty sands 080

Clean sands 072




Sands


Clays


0121 ∙ [0256 + 0336 ∙ 119861119902 + log ( )] prime 120590119907119900


26 STRESS HISTORY


10 prime 120590119901

119884119878119877 = prime 120590119907119900








10




12 028






(1+sin(emptyprime)) 119884119878119877119897119894119898119894119905 = [ ]





11

14














16 119902119905minus120590119907119900 119904119906 =

119873119896119905







18

12

119864 prime = 119863 prime

11

19 119864prime




20 119864prime

119896119904 = [119889∙(1minus1199072)]



21 053 119872119877 = (146119902119905 + 135511989111990414 + 236)244




22 119866119898119886119909 = 120588119905 ∙ 1198811199042



be estimated from

23 03 119891119904 119881119904 (119898frasl119904) = [101 ∙ log(119902119905) minus 114]167 ∙ (100 ∙ )

119902119905






13



24 0030∙(119886119888)2∙(119868119877)075

119888119907 = 11990550







25 119888119907∙120574119908 119896 =

119863prime



26

14

125

119896 asymp ( 1

) 251∙11990550








15


16








Solution




γw = 6224 pcf

17

Layer 1


17 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf

145 psi

Layer 2


17psi

γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1204 pcf 145 psi

Layer 3


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 4


CPT Material Index

7 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1121 pcf

145 psi

Layer 1


18



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi


Qtn = 35329 n = 036 lt 10 Ic = 13

Layer 2


19



fs 17 psi Fr() = 100 ∙ = 100 ∙ = 049



uo = 0 psi




σatm 145 psi

47 minus log(Qtn)]2 + [122 + log(049)]2

Ic = radic[3

Qtn = 27947 n = 041 lt 10 Ic = 14

Layer 3


20



fs 12 psi Fr() = 100 ∙ = 100 ∙ = 081







σatm 145 psi


Qtn = 947 n = 06 lt 10 Ic = 19

21

Layer 4




fs 7 psi Fr() = 100 ∙ = 100 ∙ = 060







σatm 145 psi


22

Qtn = 686 n = 064 lt 10 Ic = 19


Layer 1


Figure 11


23

Layer 2


12


24

Layer 3



25

Layer 4




Layer 1


Layer 2


26

Layer 3


Layer 4


Stress History

Layer 1

028 028


Icfrasl ) 1 + (13frasl ) 265 265


= 684 psi


primeσvo 50 psi

Layer 2

028 028



= 877 psi


primeσvo 100 psi

27

Layer 3

028 028



= 368 psi


primeσvo 164 psi

Layer 4

028 028


Icfrasl ) 1 + (19frasl ) 265 265



primeσvo 188 psi


Layer 1


Layer 2


28

Dprime 17480 psi

Eprime = = = 15890 psi 11 11




Eprime 15890 psi


053 MR = (146qt + 1355fs14 + 236)244

Layer 3

Layer 4


Layer 1


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3416 MPa = 49575 psi

fs 03




24139 kPa s

Vs = 9012 fts

29

γt 1204 pcf slug ρt = = = 37



2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000 psi s

Layer 2



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(241 MPa)053 + 1355(012 MPa)14 + 236)244 = 3415 MPa = 49562 psi

fs 03




24133 kPa s

Vs = 9012 fts

γt 1204 pcf slug ρt = = = 37


30


2 = 37 ∙ (9012 ) = 304 ∙ 106psf = 21000psi s

Layer 3



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(103 MPa)053 + 1355(008 MPa)14 + 236)244 = 1506 MPa = 21854 psi

fs 03




10343 kPa s

Vs = 8566 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8566 ) = 27 ∙ 106psf = 19000 psi s

Layer 4


31


11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(83 MPa)053 + 1355(005 MPa)14 + 236)244 = 165 MPa = 16901 psi

fs 03




8274 kPa s

Vs = 7360 fts

γt 1121 pcf slug ρt = = = 35



2 = 35 ∙ (7360 ) = 389 ∙ 106psf = 13000 psi s

32






33


34











Solution



35



Layer 1


13 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1179 pcf

145 psi

Layer 2


12 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


36

20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 4


2 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1004 pcf

145 psi

CPT Material Index

Layer 1




fs 13 psi Fr() = 100 ∙ = 100 ∙ = 043



uo = 0 psi




σatm 145 psi


37


Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 53



uo = 0 psi



38


σatm 145 psi



Layer 3




fs 16 psi Fr() = 100 ∙ = 100 ∙ = 054



uo = 0 psi



39


σatm 145 psi



Layer 4




fs 2 psi Fr() = 100 ∙ = 100 ∙ = 092



uo = 0 psi



40


σatm 145 psi




Layer 1


41


Layer 2


18

42


43

Layer 3



44

Layer 4


Figure 20



Layer 1 (sand)

45


Layer 2 (clay)

qt minus σvo


primeσvo

Where




260 psi


Layer 3 (sand)


Layer 4 (clay)

qt minus σvo


primeσvo

Where




333 psi


Stress History

46

Layer 1

028 028


Icfrasl ) 1 + (12frasl ) 265 265



primeσvo 16 psi

Layer 2

028 028


Icfrasl ) 1 + (32frasl ) 265 265



primeσvo 260 psi

Layer 3

028 028



47


primeσvo 277 psi

Layer 4

028 028




primeσvo 333 psi


Layer 1


Layer 2


Layer 3


Layer 4


48


Layer 1



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(207 MPa)053 + 1355(009 MPa)14 + 236)244 = 2816 MPa = 40873 psi

fs

03




20685 kPa s

Vs = 8412 fts

γt 1179 pcf slug ρt = = = 37



s

Layer 2

49



11 11



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(17 MPa)053 + 1355(008 MPa)14 + 236)244 = 443 MPa = 64309 psi

fs 03




17375 kPa s

Vs = 8680 fts

γt 1172 pcf slug ρt = = = 36



2 = 36 ∙ (8680 ) = 274 ∙ 106psf = 19036 psi s

Layer 3



11 11

50



053 MR = (146qt + 1355fs14 + 236)244


MR = (146(276 MPa)053 + 1355(014 MPa)14 + 236)244 = 4017 MPa = 58309 psi

fs 03




27591 kPa s

Vs = 9363 fts




2 = 38 ∙ (9363 ) = 33 ∙ 106psf = 23050 psi s

Layer 4



11 11



053 MR = (146qt + 1355fs14 + 236)244


51

MR = (146(17 MPa)053 + 1355(001 MPa)14 + 236)244 = 360 MPa = 52189 psi

fs 03




17238 kPa s

Vs = 5069 fts

γt 1004 pcf slug ρt = = = 31



s


Layer 2


12 12

Layer 4


12 12


52



Layer 1

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (7197)075

cv = = = 359 cmsec 1 sec t50

Layer 2

53

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (94392)075

cv = = = 0008 cmsec 3000 sec t50

Layer 3

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (8303)075

cv = = = 665 cmsec 06 sec t50

Layer 4

)2 ∙ (IR)075 0030 ∙ (ac 0030 ∙ (220 cm)2 ∙ (26714)075

cv = = = 087 cmsec 11 sec t50


Layer 1



Layer 2



Layer 3



54

Layer 4



55


FOUNDATIONS

31 PROCEDURE








by Equation 27





56



120782120787 minus120782120785120786120787 119956 119923 119954119950119938119961 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) ∙ (119913

) 27 119950119938119961

2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 28

119861 ℎ119904 119902119905119899119890119905











57







58




120648119959119952 = sum(120632119956119952119946119949 ∙ 119963) 29

prime 120648119959119952 = 120648119959119952 minus 119958119952 30



119954119957 = 119954119940 + 119958120784 ∙ (120783 minus 119938) 31












59

23 ℎ119904 = 28 minus

119868119888 15 32

1+( ) 24






1985 Burland 1989)



using Equation 33

33 119902119905119899119890119905 = 119902119905 minus 120590119907119900

60







Clean Sands 058 12

Silts 112 10


Intact Clays 270 4

05 minus0345 119904 119871 119902119898119886119909 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

) ∙ (119861

) 34 119898119886119909




2 119902119898119886119909 0345 1 frasl119865119878 119871 119904 = 119861 ∙ [ ∙ ∙ ( ) ] 35

119861 ℎ119904 119902119905119899119890119905




05 119904 ) 119902 = ℎ119904 ∙ 119902119905119899119890119905 ∙ (119861

minus0345 119871 ∙ ( )

11986136

61

32 EXAMPLE PROBLEMS










Bottom of layer

3 557

6 433

9 557

12 695

17 590

22 501

27 571

32 505

62


Solution



63








Layer 1


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

64

Layer 2


20psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

Layer 3


8 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1134 pcf

145 psi








65

fs 16 psi Fr() = 100 ∙ = 100 ∙ = 13





σatm 145 psi

Qtn = 703 n = 072 lt 10 Ic = 210

66






1 + ( ) 1 + ( ) 24 24

67



433 + 557 + 695 + 590 + 501 Es = = 555 tsf = 708 psi

5







B B 144

= 193 psi = 27860 psf qmax



3


68

119902119898119886119909 0345 2

1 frasl119865119878 L 119904 = 119861 ∙ [ ∙ ∙ ( ) ]

ℎ119904 119902119905119899119890119905 B


078 12311 psi 144 inches



69


41 INTRODUCTION
















70






CPT Material Index



71






421 Step 1




422 Step 2


119902119864 = 119902119905 minus 1199062 37



38 119902119887 = 119902119864 ∙ 10(0325∙119868119888minus1218)


39 119891119901 = 119902119864 ∙ 120579119875119879 ∙ 120579119879119862 ∙ 120579119877119860119879119864 ∙ 10(0732∙119868119888minus3605)


72

423 Step 3





119876119904119894119889119890 = 119891119901 ∙ 119860119904 40




= 119902119887 ∙ 119860119887 41 119876119887119886119904119890 Where


424 Step 4


119876119905119900119905119886119897 = 119876119904119894119889119890 + 119876119887119886119904119890 minus 119882119901119894119897119890 42








73

44 EXAMPLE PROBLEMS








74


75

Solution




Layer 1


76

18 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1210 pcf

145 psi

Layer 2


12psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1172 pcf

145 psi

Layer 3


20 psi γt = 624 pcf ∙ [122 + 015 ∙ ln (100 ∙ + 001)] = 1219 pcf

145 psi

CPT Material Index

Layer 1




fs 18 psi Fr() = 100 ∙ = 100 ∙ = 060



77

uo = 0 psi




σatm 145 psi


Qtn = 3593 n = 038 lt 10 Ic = 14 (ie sand)

Layer 2




fs 12 psi Fr() = 100 ∙ = 100 ∙ = 260



78

uo = 0 psi




σatm 145 psi



Layer 3




fs 20 psi Fr() = 100 ∙ = 100 ∙ = 040


79





σatm 145 psi


= 1801 n = 06 lt 10 = 15 (ie sand) Qtn Ic


Layer 1


from Figure 36

80


81

Layer 2


from Figure 37


82

Layer 3


Figure 38



Layer 1

83


Layer 2


Layer 3



Layer 3



Layer 1


fp = 2984 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙14minus3605) = 99 psi

Layer 2


fp = 468 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙29minus3605) = 211 psi

Layer 3


fp = 5000 psi ∙ 113 ∙ 111 ∙ 109 ∙ 10(0732∙15minus3605) = 202 psi

84

Axial Pile Capacity


Layer 1


Layer 2


Layer 3



85


d2 1275 in2




Layer 3



= 642 kips Qtotal


Layer L (in) qc

(psi)

fs

(psi)

Fr Ic Qtn qE

(psi)

qb

(psi)

Qbase

(lb)

fp

(psi)

Qside

(lb)

Qtotal

(kip)

1 48 3000 18 06 14 356 2984 NA NA 99 19064

2 588 500 12 26 29 117 468 NA NA 211 457122

3 240 5000 20 04 15 180 5000 9085 115932 202 58170

Total 115932 534355 642

86

REFERENCES















87

















88

















89

















90

















91
















92

















93


















94














95

APPENDIX A

List of Figures





























A-1













List of Tables





A-2


PENETRATION TESTS

A11 Introduction












interstate projects

















A-4





dashed line



A-5












details

A-6






γt Unit weight

ρt Mass Density



A-7


Table A1 Continued










G Shear modulus

E Youngs modulus


K Bulk modulus

A-8

LI



Liquefaction index




















A-9













A-10



(119902119905 minus 120590119907119900)


(119902119905 minus 120590119907119900)












A-11



(119902119905 minus 120590119907119900)120590119886119905119898 119876 = A31 prime (120590119907119900120590119886119905119898)119899




(119902119905 minus 120590119907119900) 119876 = 119899 A32

prime ) (120590119907119900



prime 120590119907119900 119899 = 0381 ∙ 119868119888 + 005 ( ) minus 015 le 10 A4 120590119886119905119898




A-12

Zone 1 119876119905119899 lt 12119890119909119901( minus14 ∙ 119865119903)

A5







A-13



Zones 8 and 9 0020)1(00030)1(0050

12

rr

tnFF

Q A6








realized







A-14









A-15

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)














A-16













A-17










A-18





prime 120590119901 = A7 033 ∙ 119902119899119890119905

prime 120590119901 = 053 ∙ ∆1199062 A8

prime 120590119901 = 060 ∙ 119902119864 A9








A-19

0

2

4

6

8

10

12

14

16

18

20

00 02 04 06 08 10 12D

ep

th (

m)


sand (SP)

Blodgett

Park Ridge

Deerfieldsoft clay

ClayLayerof Study






A-20

10

11

12

13

14

15

16

17

18

19

20

0 100 200 300 400 500 600 700 800 900 1000

De

pth

(m


Consols

033 qnet

053 Du2

060 qE

svo








A-21


(2002)


0

5

10

15

20

25

30

35

40

0 1000 2000 3000

De

pth

(m

ete

rs)


0

5

10

15

20

25

30

35

40

0 10 20 30 40 50 60


0

5

10

15

20

25

30

35

40

0 500 1000 1500

Porewater u2 (kPa)

Miscellaneous Fill

Young Bay Mud

Young Bay Mud

Clayey Sand

Miscellaneous Fill

Young Bay Mud

Young Bay Mud


A-22


0

5

10

15

20

25

30

0 100 200 300 400 500 600

De

pth

(m

ete

rs)


svo

033 qnet

053 Du2

060 qE

CRS

svo











A-23





shown by Figure A17

0

5

10

15

20

25

0 2000 4000 6000

Dep

th (

met

ers)


0

5

10

15

20

25

0 20 40 60 80 100


0

5

10

15

20

25

0 200 400 600



A-24

0

5

10

15

20

25

0 3 6 9D

epth

(m

eter


Q and Fchart

0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet

053 Du2

06 qE

Oed











A-25



58





0

5

10

15

0 500 1000 1500 2000

Dep

th (

met

ers)

Cone Tip Resistance

qt (kPa)

0

5

10

15

0 10 20 30 40 50 60

Sleeve Friction

fs (kPa)

0

5

10

15

-100 0 100 200 300 400

Porewater Pressure

u2 (kPa)


A-26

1

10

100

1000

01 1 10

Norm

aliz

ed

Con

e T

ip R

esis

tan

ce

Q




Sands

(zone 6)


Silt Mix(zone 4)

Clays

(zone 3)


Sensitive Clays

and Silts(zone 1)



sand(zone 8)

1

10

100

1000

-06 -04 -02 00 02 04 06 08 10 12 14

No

rmal

ized

Co

ne

Tip

Res

ista

nce

Q


Clays(zone 3)


(zone 2)


Sands(zone 6)





A-27

0

5

10

15

20

0 50 100 150 200 250 300 350 400

Dep

th (

met

ers)


06 qeff svo

033qnet Consols

054 Du2







(units of kPa)

prime = A11 120590119901 033 ∙ (119902119899119890119905)090



A-28





0

5

10

15

20

25

0 100 200 300 400 500 600

Dep

th (

met

ers)


svo

033qnet^090

Oed



A-29

0

5

10

15

20

0 20 40 60 80 100 120 140 160

Dep

th (

met

ers)


svo

CPT

Consols

120590119901prime = 033 119902119899119890119905

090 in kPa






120574119905 = 120574119908 ∙ [122 + 015 ∙ 119897119899 (100 lowast 119891119904120590119886119905119898 + 001)] A13


A-30



Sands





changes




A-31

120601prime(119889119890119892) = 176deg + 110deg 119897119900119892 (1199021199051) A14




(119902119905 120590119886119905119898) 1199021199051 = A151 05 prime (120590119907119900120590119886119905119898)

A-32


119902119905 1199021199051 = 05 A152 prime ) (120590119907119900



120601prime (119889119890119892) = 176deg + 110deg 119897119900119892 (119876119905119899) A16




Clays and Silts

A-33




qBQ

)tan1(tan61

1)tanexp()245(tan 2

A17


qBQ

)tan1(tan61

1)tanexp()245(tan2


1

10

100

20 25 30 35 40 45

Co

ne

Res

ista

nce

Nu

mb

er Q

(degrees)

Bq

010204060810

c = 0 = 0deg

Theory (dots)


Theory




A-34

0121 120601prime = 295deg ∙ 119861119902 ∙ [ 0256 + 0336 ∙ 119861119902 + log(119876)] A18



A19 Stress History







A-35



10

100

1000

10000

10 100 1000 10000 100000

Yie

ld S

tre

ss

s

y

(kP

a)


Blessington

General Trend

sy = 033(qt-svo)m




Clean Sands m = 072

m = 11 10 09

085

080

072

Units kPa

Fissured Clays




( ) A19 100



A-36


26 and Bq lt 01



25)652(1

2801

cIm

A20




A-37











2015)

A-38
























A-39




versus YSR or OCR


calculated from

prime TC 119904119906 = (1198721198882) ∙ (1198741198621198772)Λ ∙ 120590119907119900 A24




whereby

A-40

A25 kt

votu

N

qs

s










A-41

A-42





u

uN

us

D

D 2

A26



shear)


kE

tu

N

uqs 2

A27







119904119906(119901119890119886119896) 119878119905 = frasl A28 119904119906119903



asymp 119891119904 A29 119904119906119903






A-43

Soil Modulus








119864prime




119864prime


A-44













119864prime

119896119904 = A34 [119889middot(1minus1205842)]







A-45


Findings





Resilient modulus

Abu-Farsakh 2004

7 Louisiana sites



Mayne (2007b)



Robertson (2009)



when Ic gt 22



when Ic lt 22 then

= 003 middot 10055 Ic + 168 m

Liu et al (2016)



14 + 236)244

(all in MPa)

Casey et al (2016)








A-46

Nonlinear Modulus



119866119898119886119909 = 1198660 = 120588119905 middot 1198811199042 A35

















119866 = 119872119877119865 middot 1198660 = (1198661198660) middot 1198660 A36

A-47


1 minus (120591 )119892 119872119877119865 = (1198661198660) = frasl A37 120591119898119886119909




(1993)











A-48


``

0

01

02

03

04

05

06

07

08

09

1

000001 00001 0001 001 01 1

No

rmal

ized

GG

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

No

rmali

zed

GG

ma

x


g =1

07

05

03

02

01

0

01

02

03

04

05

06

07

08

09

1

00001 0001 001 01 1

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

0

01

02

03

04

05

06

07

08

09

1

0 1 2 3 4 5 6 7 8 9 10

No

rmali

zed

max

Shear Strain s ()

g = 1

g = 07

g = 05

g = 03

g = 02

g = 01

Reduction Factor

GGmax = 1- (max)g



A-49

0

01

02

03

04

05

06

07

08

09

1

0 01 02 03 04 05 06 07 08 09 1

Mo

du

lus

Re

du

cti

on

G

Gm

ax

or

EE

ma

x


NC SLB Sand

OC SLB Sand


Hamaoka Sand



Ham River Sand

Ticino Sand


Kaolin

Fujinomori Clay

Pietrafitta Clay

London Clay

Vallericca Clay

Pisa Clay

Pisa Clay

Pisa Clay

Pisa Clay

Kiyohoro Silty Clay

Thanet Clay

Exp g = 02

Exp g = 03

Exp g = 04

Exp g = 05

MRF = 1 - (qqmax)g












2011)

A-50










0

20

40

60

80

100

120

1 10 100 1000 10000

Pore

wat

er P

ress

ure

u2

(psi

)

Time (s)

CPTU-02E-1 z = 1560 m

uo (hydrostatic)



u2 at 50 = frac12(115+15) = 65 psi


t50 = 404 seconds


A-51









A-52













apply

A-53





A-54











consolidation cv

RemarksNotes Eqn

No

SPM

(Teh amp

Houlsby

1991) 50

2)(2450

t

Iac

Rc

v


dissipation


G = shear modulus




15-cm2 size)

A32

SCE-CSSM

(Burns amp

Mayne 2002) 50

7502 )()(0300

t

Iac Rc

v

A33

A-55





250750

max50

8111

vonet

Rq

GI

s A38








Cabal (2015)





consolidation

A-56

A40 D

ck wv







approximated by

251

50 (sec)251

1)(

tscmkh

A41

A-57



A-58

APPENDIX B

List of Figures





























B-1













List of Tables





B-2


B11 Introduction











Figure B1





B-3





al 1983)







settlements



















Schmertmann (1978)




See Figure A2


B-4




















K0 and sB)

beneath the base


2006)





beneath footing











sands


]07


Table B1 Continued



B-5





B-6



B-7


















measurements

B-8

B

D)

L

B4060(3501320kc












Cone tip resistance

measured by

mechanical CPT

Trofimenkov (1974)

Approximation

qult asymp (qc33)09

(Assuming FS = 3)


Mechanical CPT

with qc and qult in

kgcm2

Schmertmann (1978)



Cone tip resistance

measured by

mechanical CPT

Tand et al (1986)





Mix of data from

mechanical CPT qc

and electric CPT qc

LCPC Method


Footing a surface


where kc = 032

Embedment case


B-9



B-10








119902∙119861∙119868∙(1minus1205842) 119904 = B1

119864119904




















B-11















B-12











119939119943 119956 119954119938119953119953119949119946119942119941 = 119938119943 (119913

) B2

B-13





119956 119954119938119953119953119949119946119942119941 = 119955119956radic

119913 B3













B-14


footings

B-15
















B-16





Shapes and Sizes

ReferencesSource

College Station








Lehane (2008)








Amar et al (1998)


Durbin South Africa









B-17

Table B3 Continued


ReferencesSource







Brand et al (1972)







Tand et al (1986)


Tand et al (1986)


Tand et al (1986)


Tand et al (1986)








B-18









119954119940 119913

B-19















B-20



B-21

















B-22






= 119925119940 ∙ 119956119958 B5 119954119958119949119957





prime 119956119958 = 120782 120784120784120648119953 B6



prime 120648119953 = 120782 120785120785(119954119957 minus 120648119959119952) B7




B-23


















B-24



B-25






made in Equation B9

120782120787 119956 119954 = 119945119956 ∙ 119954119957119951119942119957 ∙ (119913

) lt 119954119940119938119953119938119940119946119957119962 B9






B-26







B-27

















B-28







119920119940 = radic[120785 120786120789 minus 119949119952119944119928119957119951]120784 + [120783 120784120784 + 119949119952119944119917119955]120784 B10

B-29




(119954119957minus120648119959119952)120648119938119957119950 = 119951 B11 119928119957119951 prime (120648119959119952120648119938119957119950)

119943119956 119917119955() = 120783120782120782 ∙ B12 (119954119957minus120648119959119952)





B-30



prime 120648119959119952 119951 = 120782 120785120790120783 ∙ 119920119940 + 120782 120782120787 ( ) minus 120782 120783120787 le 120783 120782 B13 120648119938119957119950




pressures)






120784120785 119945119956 = 120784 120790 minus 120783120787 B14

119920119940 120783+( ) 120784120786

B-31







)120783120787] B15

120783+(119920119940frasl120784120786




B-32








= (119912119913)120782120785120786120787 119920119912119913 B16

B-33











Footings

Mean B

(m)

Max B

(m)

Min B

(m)

Mean A

(m)

Max A

(m)

Min A

(m)

Mayne et al (2012) 32 149 609 046 149 609 046

Lehane (2011) 8 041 027 060 041 060 027

Gifford et al (1987) 17 846 1588 527 1591 3596 701

Jeyapalan amp Boehm

(1986)

13 1022 2892 400 1671 3049 640

B-34

Schmertmann

(1970)

31 945 5608 061 1288 8668 061

Papadopoulos

(1992)

29 870 3600 100 1300 7290 100

Total 130 671 5608 027 1012 8668 027



119954∙119913∙119920119912119913∙(120783minus120642120784) 119956 = B17

119916119956




)120783120787] ∙ [ ] B18

120783+(119920119940frasl120784120786 119913

B18 Conclusions









B-35



minus120782120785120786120787 119956 119912 119954119943119952119952119957119946119951119944 = 119945119956 ∙ 119954119957119951119942119957 ∙ radic ∙ ( ) lt 119954119950119938119961 B19

119913 119913






B-36

APPENDIX C

List of Figures




C-11







List of Tables




C-1


PENETRATION TESTS

C11 Introduction























C-2

119928119956 = 119943119953 middot 119912119956 C1







C2

119876119887 = 119902119887 middot 119860119887 C2


C-3






obtain

119891119901 = 120572 middot 119904119906 C3




C-4



C-5

Reference

Source


Tomlinson

(1957)

2716801102


concrete timber)

Note su in ksf

Tomlinson

(1957)

2616201102

uu ssConcrete piles

Note su in ksf

McClelland

(1974)





Note su in ksf

Semple

(1980) )1(511

)1(1

OCR


pile load tests

American

Petroleum

Institute (API

1981)





Tomlinson

(1986)

1 α = 55 (su)-091

2 α = 785 (su)-102

3 α = 605 (su)-100


1 Driven piles

2 Bored piles


lt 10middotd


than Gulf of Mexico

C-6



Kulhawy amp

Jackson

(1989)


shaft foundations

Table C1 Continued





Kulhawy amp

Jackson

(1989)

OCR

OCR

sin50

tan)sin1( sin


insensitive clays

Nowacki et al

(1996)




Kolk and van

der Velde

(1996)



L = pile length


UU lab tests


in clays

C-7




Knappett amp

Craig (2012) α = 055 ∙ 02 40

( ) (LfraslD)


minus03


grained soils



Karlsrud et al

(2005 NGI

Method)


Fleming et al


σvo NC σvo






clay

Brown et al

(2010)





Piles

Table C1 Continued

C-8

OCRK )sin1(0

Karlsrud

(2012)





prime = 120573 C4 119891119901 middot 120590119907119900






Poulos amp Davis

(1980)


Table C2 Continued

C-9

Kulhawy amp Jackson

(1989)


clays



Fleming et al

(2009)



material


Mayne and Niazi

(2017)
















given by


C-10














C-11






1977)



C8



asymp 077(120601prime75deg) 119873119902 C9









prime prime C10 119902119887 = 119873119902 ∙ 120590119907119900 ∙ 119891119909





C-12


therefore

119902119887 = 933 middot 119904119906 C11










C-13


Qside = S (fp DAs)



Qbase = qb Ab




(Scaled Pile)


type qt fs and u2)















C-14










data

Additional

parameters

needed


clays

Eslami and

Fellenius

(1997)

Fellenius

(2009)

qt fs u2

KTRI = Kajima

Technical

Research

Institute

all types Sands

mixed clays

Takesue et al

(1998)


on end bearing

resistance

Imperial College

Procedure (ICP)


(1997 PhD)

Jardine et al

(2005)


angle (δ) from

ring shear tests

Correlated to

mean grain size

(D50)

C-15


(2005)


angle (δ)

correlated to PI

Table C3 Continued

NGI Method

(Norwegian

Geotechnical

Institute)


(2005)

qt DR from qt

Clays a Alpha

method

(Karlsrud et al

2005 2012)

b Beta

method

(Karlsrud

2012)

qt for su

qt for

OCR

PI = plasticity

index ()

PI = plasticity

index ()


(2005)

qt

Clays Van Dijk and

Kolk (2011)

qt


(2005)


related to amount

of plugging

C-16

Purdue LFRD Drilled

shafts

Sands Basu amp

Salgado

(2012)

qt LRFD = load

resistance

factored design




jacked and driven

piles

UWA (Univ of

Western

Ra = pile

roughness


(2012)


angle (δ) from

ring shear tests

and also method

without δ

HKU Method

(Hong Kong

Univ)

OE and CE

(end

resistance

only)

Sands Yu and Yang

(2012)

qt End bearing only

PLR = plug length

ratio

Note PLR can be

estimated from

OE inner diam

Table C3 Continued


C-17











119902119864 = 119902119905 minus 1199062 C12




C-18



C-19






scheme




C-20









C-21





(2016)


capacity

C-22















Ps

= Sh

aft

Load

sL

t

tEd

IPw


)]1)((5ln[

)(

)1(1

1

2 vdL

dLFI

E

ECT

211

IF

P

P CT

t

b

Base Load = Pb


Ground Surface


Rigid Pile Response


L = length








21

IF

P

P CT

t

b

z = depth

= Poissons ratio

Tension

Compression

C-23


loading

C-24





C-25

Structure Bookmarks


TABLE OF CONTENTS

LIST OF TABLES





REFERENCES

APPENDIX A

APPENDIX B

APPENDIX C

cone penetration test design guide for state geotechnical

Documents